The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.
Through these waves, you can know that there is beyond the clouds.
Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.
Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.
It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.
Explanation:Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.
Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.
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A 2.0m long pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the swing?
Answer:
[tex]v=1.92m/s[/tex]
Explanation:
Given data
Length h=2.0m
Angle α=25°
To find
Speed of bob
Solution
From conservation of energy we know that:
[tex]P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\v^{2}=\frac{gh}{0.5}\\ v=\sqrt{\frac{gh}{0.5}}\\ v=\sqrt{\frac{(9.8m/s^{2} )(2.0-2.0Cos(25^{o} ))}{0.5}}\\v=1.92m/s[/tex]
Given values:
Length, h = 2.0 mAngle, α = 25°As we know,
The conservation of energy:
→ [tex]Potential \ energy = Kinetic \ energy[/tex]
or,
→ [tex]mgh = \frac{1}{2} mv^2[/tex]
or,
→ [tex]v^2 = \frac{gh}{0.5}[/tex]
[tex]= \sqrt{\frac{gh}{0.5} }[/tex]
By substituting the values, we get
[tex]= \sqrt{\frac{(9.8)(2.0-2.0 Cos (25^{\circ}))}{0.5} }[/tex]
[tex]= 1.92 \ m/s[/tex]
Thus the above answer is right.
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A book slides off a horizontal tabletop. As it leaves the table’s edge, the book has a horizontal velocity of magnitude v0. The book strikes the floor in time t. If the initial velocity of the book is doubled to 2v0, what happens to (a) the time the book is in the air, (b) the horizontal distance the book travels while it is in the air, and (c) the speed of the book just before it reaches the floor? In particular, does each of these quantities stay the same, double, or change in another way? Explain.
Answer:
(a) The time the book is in the air stays the same.
(b) The horizontal distance the book travels doubles.
(c) The speed of the book just before it reaches the floor increases but not doubles.
Explanation:
The following kinematics equations will be used to solve this question:
[tex]v_y = v_{y_0} + a_yt\\y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)[/tex]
(a) Initially, the y-component of the velocity is zero. So, the x-component of the velocity is doubled.
We will use the second equation for both cases:
[tex]0 - y_0 = v_{y_0}t - \frac{1}{2}(g)t^2 = 0 - \frac{1}{2}gt^2\\0 - y_0 = v_{y_0}t_2 -\frac{1}{2}gt_2^2 = 0 - \frac{1}{2}gt_2^2\\t = t_2[/tex]
Since, the initial velocity in the y-direction is zero, and the height of the table is constant. The time it takes from the edge of the table to the floor is the same.
(b) For the horizontal distance the book travels, we should use the second equation again, and keep in mind that the acceleration in the x-direction is zero.
[tex]x - 0 = v_{x_0}t + \frac{1}{2}a_xt^2 = v_0t\\x_2 = 2v_0t_2 = 2v_0t[/tex]
Hence, the horizontal distance doubles.
(c) The vertical velocity does not change, since the initial velocity in the y-direction is zero. The horizontal velocity does not change along the motion, since the acceleration in the x-direction is zero. However, since the initial velocity in the x-direction doubles, the final velocity in the x-direction doubles as well.
[tex]v_x = v_{x_0} + a_xt = v_{x_0}[/tex]
However, this does not mean that final speed of the book will double. Because the speed of the object is calculated as follows:
[tex]v_{\rm final_1} = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + v_y^2}\\v_{\rm final_2} = \sqrt{v_{x_2}^2 + v_y^2} = \sqrt{(2v_0)^2 + v_y^2}[/tex]
As can be seen from above, the final speed increases but not doubles.
Doubling the initial horizontal velocity of a book in projectile motion would result in the same time in the air, double the distance covered, and a higher (but not doubled) speed when it hits the floor.
Explanation:This question explores the concepts of projectile motion in physics. When we double the initial horizontal velocity of the book from v0 to 2v0, the following changes occur:
Time the book is in the air: This will stay the same. The time a projectile is in the air is determined by the vertical component of its motion alone. Thus, the horizontal velocity does not affect the time.Horizontal distance the book travels while it is in the air: This will double. The horizontal distance (d) a projectile travels is determined by the horizontal velocity (v) and the time it spends in the air (t). So, d=v*t. If we double the initial horizontal velocity, the distance will double as well.Speed of the book just before it reaches the floor: This will increase. The speed a projectile hits the floor with is determined by both its horizontal and vertical speeds. Doubling the initial horizontal velocity will result in a greater overall speed when it hits the floor, but not necessarily double the original speed.Learn more about Projectile Motion here:https://brainly.com/question/20627626
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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then runs out of fuel. Ignore any air resistance effects. How long is the rocket in the air before hitting the ground?
Answer:
T = 295.57 s
Explanation:
given,
mass of the rocket = 200 Kg
mass of the fuel = 100 Kg
acceleration = 35 m/s²
time, t = 35 s
time taken by the rocket to hit the ground, = ?
Final speed of the rocket when fuel is empty
using equation of motion
v = u + a t
v = 0 + 35 x 35
v = 1225 m/s
height of the rocket where fuel is empty
v² = u² + 2 a s
1225² = 0 + 2 x 35 x h₁
h₁ = 21437.5 m
After 35 s the rocket will be moving upward till the final velocity becomes zero.
Now, using equation of motion to find the height after 35 s
v² = u² + 2 g h₂
0² = 1225² + 2 x (-9.8) h₂
h₂ = 76562.5 m
total height = h₁ + h₂
= 76562.5 m + 21437.5 m = 98000 m
now, time taken by before the rocket hit the ground
using equation of motion
[tex]s = u t +\dfrac{1}{2}at^2[/tex]
[tex]-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2[/tex]
negative sign is used because the distance travel by the rocket is downward.
4.9 t² - 1225 t - 13500 = 0
[tex]t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}[/tex]
t = 260.57 s
neglecting the negative sign
total time the rocket was in air
T = t₁ + t₂
T = 35 + 260.57
T = 295.57 s
Time for which rocket was in air is equal to 295.57 s.
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each section of the string moves with simple harmonic motion at a frequency of 8 Hz. Find the power propagated along the string.
Answer:
Power of the string wave will be equal to 5.464 watt
Explanation:
We have given mass per unit length is 0.050 kg/m
Tension in the string T = 60 N
Amplitude of the wave A = 5 cm = 0.05 m
Frequency f = 8 Hz
So angular frequency [tex]\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec[/tex]
Velocity of the string wave is equal to [tex]v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec[/tex]
Power of wave propagation is equal to [tex]P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt[/tex]
So power of the wave will be equal to 5.464 watt
A proton is observed to have an instantaneous acceleration of 10 × 1011 m/s2. What is the magnitude of the electric field at the proton's location?
To solve this problem we will apply the concept of Newton's second law with which we will obtain the strength of the proton. We know the mass and the acceleration is given in the statement. Subsequently said Force by equilibrium can be matched the electrostatic force of Coulomb, defined as the product between the charge and the electric field. Our values are
[tex]m = 1.67*10^{-27}kg[/tex]
[tex]a = 10*10^{11} m/s^2[/tex]
Applying the Newton's second law,
[tex]F = ma[/tex]
[tex]F = (1.67*10^{-27}kg)( 10*10^{11} m/s^2)[/tex]
[tex]F = 1.67*10^{-15}N[/tex]
By the Coulomb's equation for electrostatic Force we have that
[tex]F = qE[/tex]
Remember that the charge of a proton is [tex]1.6*10^{-19}C[/tex]
Replacing we have,
[tex]1.67*10^{-15} = 1.6*10^{-19} E[/tex]
[tex]E = 10437.5 N/C[/tex]
Therefore the magnitude of the electric field at the proton's location is [tex]10437.5 N/C[/tex]
The magnitude of the electric field at the proton's location is 10,437.5 N/C.
The given parameters:
Acceleration of the proton, a = 10 x 10¹¹ m/s²Mass of proton, m = 1.67 x 10⁻²⁷ kgThe magnitude of the electric field at the proton's location is calculated as follows;
F = ma
F = qE
qE = ma
[tex]E = \frac{ma}{q} \\\\E = \frac{1.67 \times 10^{-27} \times 10\times 10^{11}}{1.6 \times 10^{-19}} \\\\E = 10,437.5 \ N/C[/tex]
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a 282 kg bumper car moving +3.70 m/s collides with a 155 kg bumper car moving -1.38 m/s. afterwards, the 282 kg car moves at +1.10 m/s. find the velocity of 155 kg car afterwards.(unit=m/s) Please help me !! I have been stuck on this question for 1 hour
Answer: 3.35 m/s
Explanation:
This problem is related to the Conservation of Momentum principle, which states the following:
"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision will be the same as the total momentum of these same two objects after the collision".
Of course, the momentum of each object may change after the collision, however, the total momentum of the system does not change. In addition, according to this law it is established that the initial momentum [tex]p_{o}[/tex] must be equal to the final momentum [tex]p_{f}[/tex]:
[tex]p_{o}=p_{f}[/tex] (1)
Where:
[tex]p_{o}=m_{1}V_{1}+m_{2}V_{2}[/tex] (2)
[tex]p_{f}=m_{1}U_{1}+m_{2}U_{2}[/tex] (3)
Being:
[tex]m_{1}=282 kg[/tex] the mass of the first car
[tex]V_{1}=3.70 m/s[/tex] the initial velocity of the first car
[tex]m_{2}=155 kg[/tex] the mass of the second car
[tex]V_{2}=-1.38 m/s[/tex] the initial velocity of the second car
[tex]U_{1}=1.10 m/s[/tex] the final velocity of the first car
[tex]U_{2}[/tex] the final velocity of the second car
Substituting (2) and (3) in (1):
[tex]m_{1}V_{1}+m_{2}V_{2}=m_{1}U_{1}+m_{2}U_{2}[/tex] (4)
Isolating [tex]U_{2}[/tex]:
[tex]U_{2}=\frac{m_{1}(V_{1}-U_{1})+m_{2}V_{2}}{m_{2}}[/tex] (5)
Solving:
[tex]U_{2}=\frac{282 kg(3.70 m/s-1.10 m/s)+(155 kg)(-1.38 m/s)}{155 kg}[/tex] (6)
Finally:
[tex]U_{2}=3.35 m/s[/tex]
Answer:
3.35
Explanation:
If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be changed. That is, what is the ratio of the new force to the old force?
Answer:
4
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
[tex]m_1[/tex] = Mass of Earth
[tex]m_2[/tex] = Mass of Moon
r = Distance between Earth and Moon
Old gravitational force
[tex]F_o=\dfrac{Gm_1m_2}{r^2}[/tex]
New gravitational force
[tex]F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}[/tex]
Dividing the equations
[tex]\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4[/tex]
The ratio is [tex]\dfrac{F_n}{F_o}=4[/tex]
The new force would be 4 times the old force
If the distance between the Earth and Moon were halved, the force of gravity between them would be quadrupled.
Explanation:According to Newton's law of universal gravitation, the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, this relationship is expressed as:
F= G*m*M/r^2
[tex]r^{2}[/tex]If the distance between the Earth and Moon were halved, the force of gravity between them would be quadrupled. This is because the force of gravity is inversely proportional to the square of the distance between two objects. So, if the distance is divided by 2, the new force would be multiplied by (2^2) = 4.
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How many times does a human heart beat during a person’s lifetime? How many gallons of blood does it pump? (Estimate that the heart pumps 50 cm3 of blood with each beat.)
Answer: The heart pumps 124.2 billion cm³ of blood in a lifetime
Explanation:
as an adult the pulse rate average must be around 72 beats per minute.
The heart beats about 103,680 times in a day.
There are 365 days in a year
number of heart beat in a year = 365 days x 103,680 = 37,843,200 beats in a year
For every the heart pumps 50cm³ of blood,
Hence,
Amount of blood pump in a year = 50 x 37,843,200 = 1,892,160,000cm³ of blood pumped in a year.
Using the estimated lifespan average an individual is 69 years
So in a life time,
The human heart pumps = 1,892,160,000 x 69 years = 124,200,000,000
If the heart pumps 50cm³ of blood per beat, the heart pumps a total of 130,559,040,000 cm³ (130.6 billion cm³) of blood in a LIFETIME.
The human heart beats approximately 108,000 times per day, amounting to nearly 3 billion beats in a 75-year lifespan and pumps about 2.6 million gallons of blood. To estimate the volume in cubic meters, we use the flow rate of 5 L/min over a 75-year period and convert liters to cubic meters.
Estimating Human Heart Beat and Blood Volume Pumped Over a Lifetime
The vital importance of the heart is evident in its tireless work throughout a person's life. Assuming an average heart rate of 75 beats per minute, we can estimate that a human heart beats about 108,000 times in one day, which amounts to more than 39 million times in one year, and nearly 3 billion times during a 75-year lifespan. When it comes to the volume of blood pumped, with each contraction pumping approximately 70 mL of blood, the heart pumps roughly 5.25 liters of blood per minute. This translates to about 14,000 liters per day, and over a year, the heart would pump approximately 10,000,000 liters, or roughly 2.6 million gallons of blood through an extensive network of vessels.
A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis)
Incomplete question as we have not told to find any quantity so I have chosen some quantities to find.So complete question is here
A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations.
Find (a) the force constant of the spring
(b) the frequency of the oscillations
(c) the maximum speed of the object.
Answer:
(a) [tex]k=100N/m[/tex]
(b) [tex]f=1.126Hz[/tex]
(c) [tex]v_{max}=1.41m/s[/tex]
Explanation:
Given data
Mass of object m=2.00 kg
Horizontal Force F=20.0 N
Distance of the object from equilibrium A=0.2 m
To find
(a) Force constant of the spring k
(b) Frequency F of the oscillations
(c) The Maximum Speed V of the object
Solution
For (a) force constant of the spring k
From Hooke's Law we know that:
[tex]F=kx\\k=F/x\\where\\x=A(amplitude)\\So\\k=F/A\\k=(20N/0.2000m)\\k=100N/m[/tex]
For (b) frequency F of the oscillations
The frequency of the motion for an object in simple harmonic motion is expressed as:
[tex]f=\frac{1}{2\pi } \sqrt{\frac{k}{m} }\\ f=\frac{1}{2\pi } \sqrt{\frac{100N/m}{2.0kg} }\\f=1.126Hz[/tex]
For (c) maximum speed V of the object
For an object in simple harmonic motion the maximum values of the magnitude velocity is given as:
[tex]v_{max}=wA\\ where\\w=\sqrt{\frac{k}{m} }\\ so\\v_{max}=\sqrt{\frac{k}{m} }A\\v_{max}=\sqrt{\frac{100N/m}{2.0kg} }(0.200m)\\v_{max}=1.41m/s[/tex]
a projectile is fired in the earths gravitational field with a horizontal velocity in v = 9.00m/s. how far does it go in the horizontal direction in .550 s?
Answer:
Distance traveled by projectile in horizontal direction will be 4.95 m
Explanation:
We have given horizontal velocity of the projectile v = 9 m/sec
We have to find the distance traveled by projectile in 0.550 sec in horizontal direction.
We know that distance is equal to multiplication of speed and time
So distance traveled in horizontal direction will be equal to [tex]d=v\times t=9\times 0.550=4.95m[/tex]
So distance traveled by projectile in horizontal direction will be 4.95 m
Answer:
4.95 m
Explanation:
Horizontal velocity of projectile, u = 9 m/s
time, t = 0.550 s
The horizontal distance, d = horizontal velocity x time
d = 9 x 0.55
d = 4.95 m
Thus, the horizontal distance traveled is 4.95 m.
A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude of the electric field is 1.36 105 N/C.
(a) What is the speed of the proton after it has traveled 3.00 cm?
(b) What is the speed of the proton after it has traveled 30.0 cm?
Answer:
a) 8.83*10⁵ m/s b) 2.80*10⁶ m/s
Explanation:
a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.
By definition, the force due to the electric field is as follows:
F = q*E = e*E (1)
where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.
According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:
F = mp*a (2)
From (1) and(2), being left sides equal, right sides must be equal too:
[tex]F = e*E = mp*a[/tex]
Solving for a:
[tex]a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2[/tex]
⇒ a = 1.3*10¹³ m/s²
As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:
[tex]vf^{2} -vo^{2} = 2*a*x[/tex]
Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and x = 0.03 m, we can find vf as follows:
[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s[/tex]
⇒ vf = 8.83*10⁵ m/s
b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:
[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s[/tex]
⇒ vf = 2.80*10⁶ m/s
To find the speed of the proton after traveling a certain distance in a uniform electric field, we can use the equations of motion and the equation for force. By plugging in the known values and solving for acceleration, we can then use the equation for speed to find the desired velocity.
To find the speed of the proton, we can use the equation:
F = ma
where F is the force, m is the mass, and a is the acceleration.
The force on the proton is given by:
F = qE
where q is the charge of the proton and E is the electric field strength.
Plugging in the values, we get:
qE = ma
Solving for acceleration, we have:
a = qE/m
For the first part of the question, with a distance of 3.00 cm, we can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is 0), a is the acceleration, and s is the distance.
Plugging in the known values, we have:
v^2 = 0 + 2 * (qE/m) * s
Simplifying:
v = sqrt(2 * (qE/m) * s)
For the second part of the question, with a distance of 30.0 cm, we can use the same equation:
v^2 = u^2 + 2as
Plugging in the known values:
v^2 = 0 + 2 * (qE/m) * s
Simplifying:
v = sqrt(2 * (qE/m) * s)
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A person riding in a boat observes that the sunlight reflected by the water is polarized parallel to the surface of the water. The person is wearing polarized sunglasses with the polarization axis vertical.
If the wearer leans at an angle of 19.5 degrees to the vertical, what fraction of the reflected light intensity will pass through the sunglasses?
The angle of incidence, when there is perfect transmission is the polarization angle. The expression for the polarization is
[tex]\theta = 90-\phi[/tex]
Where
[tex]\theta[/tex] = Polarization angle
[tex]\phi[/tex] = Angle with the vertical axis
We have that the angle is
[tex]\theta = 90-19.5[/tex]
[tex]\theta = 70.5\°[/tex]
The ratio of the intensities depends on the cosine of the polarization angle. The polarization angle found from the wearer’s leaning angle can be used to find the fraction of the reflected ray intensity that will pass through the sunglasses.
Applying the Malus Law we have that
[tex]\frac{I}{I_0} = cos^2 \theta[/tex]
Here,
[tex]I[/tex] = Final intensity
[tex]I_0[/tex] = Initial intensity
Replacing we have,
[tex]\frac{I}{I_0} = cos^2 (70.5)[/tex]
[tex]\frac{I}{I_0} = 0.11[/tex]
Therefore the fraction of the reflected light intensity which passes through the sunglasses is 0.11
By applying Malus's Law, a person leaning at 19.5° while wearing polarized sunglasses with a vertical axis lets approximately 88.9% of the reflected, polarized light pass through.
A person riding in a boat observes that the sunlight reflected by the water is polarized parallel to the surface of the water. The person is wearing polarized sunglasses with the polarization axis vertical. If the wearer leans at an angle of 19.5 degrees to the vertical, we can determine the fraction of the reflected light intensity that will pass through the sunglasses using Malus's Law.Malus's Law states that the transmitted light intensity I through a polarizing filter is given by:I = Io cos²(θ)
where I is the transmitted intensity, Io is the initial light intensity, and θ is the angle between the light's polarization direction and the transmission axis of the filter.Since the observer leans at an angle of 19.5°, the angle between the polarized light and the vertical axis of the sunglasses is 19.5°. Applying Malus’s Law:I = Io cos²(19.5°)
First, compute cos(19.5°):cos(19.5°) ≈ 0.943
Then, square this value:(0.943)² ≈ 0.889
Therefore, the fraction of the reflected light intensity that will pass through the sunglasses is approximately 0.889 or 88.9%.
After fixing a flat tire on a bicycle you give the wheel a spin. If its initial angular speed was 6.36 rad/s and it rotated 14.7 revolutions before coming to rest, what was its average angular acceleration (assuming that the angular acceleration is constant)
To solve this problem we will apply the concepts related to the cinematic equations of angular motion. On these equations, angular acceleration is defined as the squared difference of angular velocity over twice the radial displacement. This is mathematically:
[tex]\alpha = \frac{\omega^2-\omega_0^2}{2\theta}[/tex]
Our values are,
[tex]\text{Initial angular velocity} = \omega_0 =6.36 rad/s[/tex]
[tex]\text{Final angular velocity} = \omega =0[/tex]
[tex]\text{Angular displacement} = \theta = 14.7rev = 29.4\pi rad[/tex]
Replacing,
[tex]\alpha = \frac{- 6.36^2}{29.4\pi}[/tex]
[tex]\alpha = -0.43rad/s^2[/tex]
Therefore the angular acceleration is [tex]-0.43rad/s^2[/tex]
A partially divided tank contains two immiscible fluids oil (oil = 898 kg/m3 ) and water ( = 998 kg/m3 ). What is the height, h, of the oil column above the tank top? (8.0 cm)
Answer:
The height of the oil column above the tank top is 8 cm.
Explanation:
By applying Bernoulli's equation between point A and B as shown in the attached diagram
[tex]P_{atm}+\rho_{oil}g(h+0.12)=P_{atm}+\rho_{water}g(0.06+0.12)[/tex]
Here
P_atm is the atmospheric pressure.ρ_water=998 kg/m3ρ_oil=898 kg/m3g=9.8 m/s2[tex]P_{atm}+\rho_{oil}g(h+0.12)=P_{atm}+\rho_{water}g(0.06+0.12)\\898\times (h+0.12)=998 \times (0.06+0.12)\\(h+0.12)=\frac{998}{898} \times (0.18)\\h+0.12=0.20\\h=0.20-0.12\\h=0.08m \approx 8 cm[/tex]
So the height of the oil column above the tank top is 8 cm.
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 7.0m/s . Once free of this area, it speeds up to 12m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final Speed?
Answer:
[tex] v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s[/tex]
Explanation:
For this case we know that the initial velocity is given by:
[tex] v_i = 7 \frac{m}{s}[/tex]
The final velocity on this case is given by:
[tex] v_f = 13 \frac{m}{s}[/tex]
And we know that it takes 8 seconds to go from 7m/s to 13m/s. We can use the following kinematic formula in order to find the acceleration during the first interval:
[tex] v_f = v_i +at[/tex]
If we solve for the acceleration we got:
[tex] a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}[/tex]
So for the other traject we assume that the acceleration is constant and the train travels for 16 s. The initial velocity on this case would be 13m/s from the first interval and we can find the final velocity with the following formula:
[tex] v_f = v_ i +a t[/tex]
And if we replace we got:
[tex] v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s[/tex]
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C
To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire. The distance from the line where the electric field strength is 2000N/C is d = lambda / 2 x 10^6 m.
Explanation:To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire:
E = k * (lambda / d)
Where E is the electric field strength, k is the Coulomb's constant, lambda is the charge density of the wire, and d is the distance from the wire. In this case, since the line of charge is infinite, the charge density is simply the charge per unit length.
To solve for the distance d, we can rearrange the formula: d = k * (lambda / E)
Plugging in the given values, the distance d is:
d = (9.0 x 10^9 Nm^2/C^2) * (lambda / 2000 N/C)
So, the distance from the line where the electric field strength is 2000 N/C is d = lambda / 2 x 10^6 m.
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List two reasons for why a non-physicist might be interested in electrostatic interactions.
Answer:
all areas of knowledge that wish to understand the physical, chemical and biological process must know electrostatics
Explanation:
Electrostatic interactions, have many rare manifestations in nature, which causes many reasons to study them.
- Lightning is a very striking form of electricity
- The biological processes are governed by currents of inanes and potential differences
- The transfer of nutrients and fertilizers to plants is with ion exchange, electrostatic forces
- all modern electronics is based on electricity
- the electric charge in very dry places, creates high currents that can create fires or kill people
In summary all areas of knowledge that wish to understand the physical, chemical and biological process must know electrostatics
A non-physicist might be interested in electrostatic interactions due to their real-world applications, such as photocopiers and air filters, as well as their significance in biological systems.
Explanation:A non-physicist might be interested in electrostatic interactions for a couple of reasons. First, electrostatic interactions play a significant role in several real-world applications, such as photocopiers, laser printers, ink-jet printers, and electrostatic air filters. Understanding these interactions can help in understanding how these technologies work and how they can be improved. Second, electrostatic interactions are also important in biological systems, where they affect the interactions between molecules. This knowledge is relevant in fields like medicine and biology.
Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocity, etc.) In particular, by jumping they can raise their centers of mass the same vertical distance, H (called their "vertical leap"). The first player, Arabella, wishes to shoot over the second player, Boris, and for this she needs to be as high above Boris as possible. Arabella jumps at time t=0, and Boris jumps later, at time tR (his reaction time). Assume that Arabella has not yet reached her maximum height when Boris jumps.
Answer:
(a). D(t) = (√2gh)t - 1/2gt²
(b). D(t) = tk [(√2gh) + g(tk - 2t) / 2]
Explanation:
we would be using the equation of motion to determine the distance D
from the question;
the height of the hand raised by Boris isH-boris = h₀ + V₀ (t - tk) - 1/2g(t -tk)²
Also given is the height of Arabella raised hand is given thus;H-arabella = h₀ + V₀t - 1/2gt²
(a). the vertical displacement is given;
D(t) = H-arabella - H-boris
D(t) = h₀ + V₀t -1/2gt₂ - (h₀ + V₀(t-tk) -1/2g(t-tk)₂)
D(t) = h₀ + V₀t - 1/2gt₂ - h₀ where 0 ∠ t ∠ tk
this gives D(t) = V₀t -1/2gt²
where V₀ = √2gh
∴ D(t) = (√2gh)t - 1/2gt²
(b). We already know vertical displacement as;
D(t) = H-arabella - H-boris
D(t) = h₀ + V₀t -1/2gt₂ - (h₀ + V₀(t-tk) -1/2g(t-tk)₂)
= V₀tk - 1/2gt² + 1/2g(t -tk)²
= V₀tk + 1/2gtk² - gttk
= √2gh tk + 1/2gt² - gttk
this gives D(t) = tk [(√2gh) + g(tk - 2t) / 2]
cheers i hope this helps.
A 1.00-kmkm length of power line carries a total charge of 230 mCmC distributed uniformly over its length. Find the magnitude of the electric field 65.1 cmcm from the axis of the power line, and not near either end (staying away from the ends means you can approximate the field as that of an infinitely long wire). Express your answer with the appropriate units.
Answer:
[tex]E = 6.38\times 10^6~N/C[/tex]
Explanation:
The question states that we can approximate the line as an infinite wire. In that case, the electric field can be found by Gauss' Law.
We should draw an imaginary cylindrical surface with an arbitrary height, h, around the wire. The radius of the cylinder should be equal to 65.1 cm.
Gauss' Law:
[tex]\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]
The integral in the left-hand side is not to be taken, because we know the area of the cylinder. The enclosed charge in the right-hand side is equal to the charge of the portion of the wire inside the imaginary surface.
The charge density of the wire is
[tex]\lambda = \frac{Q}{L} = \frac{230 \times 10^{-3}}{1000} = 2.3 \times 10^{-4}[/tex]
The charge enclosed by the imaginary surface is
[tex]Q_{enc} = \lambda h = 2.3\times 10^{-4}h[/tex]
Finally, Gauss' Law yields
[tex]E2\pi rh = \frac{\lambda h}{\epsilon_0}\\E = \frac{\lambda}{2\pi \epsilon_0r} = \frac{2.3 \times 10^{-4}}{2\pi\epsilon_0(65.1\times 10^{-2})} = 6.38\times 10^6~N/C[/tex]
A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate (a) the flexural strength; and (b) the flexural modulus, assuming that no plastic deformation occurs.
The flexural strength of the ZrO2 block is 7680 lb/in^2. The flexural modulus of the ZrO2 block is 33546.74 lb/in^2.
Explanation:To calculate the flexural strength of the ZrO2 block, we need to use the formula:
Flexural strength = (3F * L) / (2 * b * h^2)
where F is the applied force, L is the distance between supports, b is the width of the block, and h is the thickness of the block. Substituting the given values, we have:
Flexural strength = (3 * 400 lb * 8 in.) / (2 * 0.50 in. * (0.25 in.)^2) = 7680 lb/in^2
To calculate the flexural modulus, we can use the formula:
Flexural modulus = (F * L^3) / (4 * b * h^3 * y)
where y is the deflection of the block. Substituting the given values, we have:
Flexural modulus = (400 lb * (4 in.)^3) / (4 * 0.50 in. * (0.25 in.)^3 * 0.037 in.) = 33546.74 lb/in^2
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A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first 2.0 s of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) 50.0 m; (ii) 100.0 m; (iii) 200.0 m?
A point charge is placed at the center of a spherical Gaussian surface. The electric flux is changed: A. if the sphere is replaced by a cube of the same volume Explain B. if the sphere is replaced by a cube of one-tenth the volume C. if the point charge is moved off center, but still inside the original sphere D. if the point charge is moved to just outside the sphere E. if a second point charge is placed just outside the sphere
Answer:
Explanation:
Point charge is Placed at the center of a spherical Gaussian surface and according to Gauss law electric flux through Gaussian surface is given by [tex]\frac{1}{\epsilon _0}[/tex] times the charge enclosed
where [tex]\epsilon _0=[/tex]permittivity of free space
So unless we take charge outside of the surface flux will not change for the Gaussian surface
So Statement A,B,C and E are false
Option D is correct because we take the charge outside of the Gaussian surface
The electric flux depends on the charge enclosed by the Gaussian surface and the location of the charges. If the volume of the surface changes, but the charge remains inside, the flux remains the same. If the charge is moved off-center or outside the surface, or if a second charge is added outside the surface, the flux will change.
Explanation:The electric flux depends on the charge enclosed by the Gaussian surface. So, if the sphere is replaced by a cube of the same volume with the charge at the center, the electric flux will not change. However, if the cube is one-tenth the volume of the original sphere, the electric flux will decrease by a factor of 10, since the charge is still at the center.
If the point charge is moved off center, but still inside the original sphere, the electric flux will not change, as long as the charge remains enclosed by the Gaussian surface.
If the point charge is moved to just outside the sphere, the electric flux will increase, as some electric field lines are now passing through the Gaussian surface.
If a second point charge is placed just outside the sphere, the electric flux through the Gaussian surface will increase even more, as the electric field lines from both charges are now passing through the surface.
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(a) If the electric field is zero in some region of space, the electric potential must also be zero in that region. a. true b. false (b) If the electric potential is uniformly zero in some region of space, the electric field must also be zero in that region. a. true b. false (c) If the electric potential is zero at a point, the electric field must also be zero at that point. a. true d. false (d) Electric field lines always point toward regions of lower potential. a. true b. false(e) The value of the electric potential can be chosen to be zero at any convenient point. a. true b. false (f) In electrostatics, the surface of a conductor is an equipotential surface. a. true b. false(g) Dielectric breakdown occurs in air when the potential is 3 times 106 V. a. true b. false
The correct options are a) a. false b) b. false c) a.True d) a.True e) a.True f) a.True g) b. false.
(a) The given statement is incorrect, An electric field is the gradient of the potential. when the Potential is constant The electric field is zero. The correct option is b. false
(b) If an electric field of space is zero, it does not imply that there is no charge . From Gauss law in its divergence form, It implies there are an equal number of opposite charges present. No statement can be said about the charges present in that region. The correct option is b. false.
(c) The statement is correct because the Intensity of an electric field is line integral of the electric potential. The correct otption is a. True
(d) Electric field lines always point from high potential to low potential. The correct option is a. True
(e) When the electric field strength is not zero, an electric potential is zero at all points on the equatorial line of the electric dipole. The correct option is a. True
(f) The statement is correct as conductors allow the free flow of charge within themselves. The correct statement is a. True
(g) It is incorrect as dielectric breakdown occurs when a charge buildup exceeds the electrical limit or dielectric strength of a material. The correct option is b. False
The options are a) false b) false c) True d) True e) True f) True g) false.
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Final answer:
The relationship between electric fields and electric potential is complex. Zero electric field does not guarantee zero electric potential, and equipotential surfaces arise when there is no change in potential, forming naturally around conductors in electrostatic conditions.
Explanation:
Addressing the statements related to electric fields and electric potential:
False: If the electric field is zero in some region of space, it does not imply that the electric potential is also zero; the potential could have a nonzero constant value.
True: If the electric potential is uniformly zero, the electric field must also be zero since a nonzero field would indicate a change in potential.
False: Zero electric potential at a point does not necessarily mean that the electric field is zero at that point.
True: Electric field lines always point from regions of higher to lower potential.
True: The value of the electric potential can be chosen to be zero at any convenient reference point.
True: In electrostatics, the surface of a conductor is an equipotential surface because the electric field within a conductor must be zero.
False: The breakdown voltage of air varies depending on conditions such as air density and humidity, and is generally accepted to be approximately 3×10⁶ V/m but the exact value can differ.
Which of the following liquids will have the lowest freezing point/ A) pure h20 B)aqueous KF(0.50 m) C)aqueous sucrose (0.60 m) D)aqueous glucose (0.60 m) E) aqueous Fel3 (0.24 m)
The aqueous solution with the highest concentration of particles will have the lowest freezing point.
Explanation:The freezing point of a solution is determined by the concentration of solute particles in the solution. The greater the concentration of particles, the lower the freezing point. In this case, the aqueous solution with the lowest freezing point will have the highest concentration of particles.
Out of the options given, the aqueous solution with the highest concentration of particles is 0.60 m aqueous sucrose. Therefore, aqueous sucrose (0.60 m) will have the lowest freezing point.
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If the focal length of a reflection telescope is 200 cm and the focal length of the eyepiece lens is 0.25 cm, what is the magnifying power of the telescope?
Answer:
800
Explanation:
Focal length of telescope, F = 200cm
Focal length of eyepiece, f = 0.25
The magnifying power of a telescope is given as the ratio of the focal length of the objective of the telescope to the focal length of the lens. Mathematically:
M = F/f
Therefore, when F = 200cm and f = 0.25cm:
M = 200/0.25
M = 800
Answer:
-48cm
Explanation:
the following data are given
focal length of telescope=200cm,
focal length of the eyepiece=0.25cm
From the genera formula used to find the magnifying power which is expressed as
[tex]M=-\frac{fx_{o}}{f_{e}}[1+\frac{f_{e}}{d}][/tex]
where
[tex]f_{e} = focal length of thr eye piece\\ f_{o} =focal length of the telescope\\[/tex]
and d=least distance of distinct vision=25cm
if we substitute values into the formula, we arrive at
[tex]M=-\frac{fx_{o}}{f_{e}}[1+\frac{f_{e}}{d}]\\M=-\frac{200cm}{0.25cm}[1+\frac{0.25cm}{25cm}]\\M=-800cm[1+0.01]\\M=-800cm(1.01)\\M=-808cm \\M=-808cm[/tex]
hence from the answer, we can conclude that the magnifying power of the telescope is -808cm
A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a stationary car that has a perfectly elastic spring bumper. What is the final kinetic energy of the two car system?
To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,
[tex]KE_i = KE_f[/tex]
[tex]KE_f = \frac{1}{2} mv^2[/tex]
Here,
m = Mass
V = Velocity
Replacing,
[tex]KE_f = \frac{1}{2} (12000)(11)^2[/tex]
[tex]KE_f = 72600J[/tex]
Therefore the final kinetic energy of the two car system is 72.6kJ
The final kinetic energy of two car system is 72,600 Joules.
To understand more, check below explanation.
Energy conservation:The energy is neither be created nor be destroyed only transfer from one form to another form.
So that,
Initial kinetic energy = Final kinetic energy
It is given that, mass,m = 1200kg , speed, v = 11m/s.
As we know that,
Kinetic energy[tex]=\frac{1}{2}*m*v^{2}[/tex]
Substute above values in above formula.
[tex]K.E=\frac{1}{2}*1200*(11)^{2} \\\\K.E=600*121=72,600J[/tex]
Hence, the final kinetic energy of two car system is 72,600 Joules.
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A single-turn rectangular wire loop measures 6.00 cm wide by 10.0 cm long. The loop carries a current of 3.00 A. The loop is in a uniform magnetic field with B = 4.00 times 10^-3 T. Taking torques about an axis, parallel to either side of the rectangular loop, that maximizes the torque, what is the magnitude of the torque exerted by the field on the loop if the direction of the magnetic field is described as the following? a) parallel to the short sides of the loop N middot m b) parallel to the long sides of the loop N middot m c) perpendicular to the plane of the loop N middot m
Answer:
Explanation:
Single turn implies that N=1
The rectangle is 6cm (0.06m) wide and 10 cm (0.1m) long.
Current in loop is 3A
Magnetic Field B= 4×10^-3T
To take the maximum torque then, sinθ=1
Then
τ=NiABsinθ
Where N is number of turns, and in this case N is 1
i is current in coil and it this case it is 3Amps.
A is area of of coil
And in this case it is a rectangle
Area of rectangle is Lenght × breadth
A= 0.06×0.1= 0.006m^2
And B is magnetic field given as B=4×10^-3T
And θ is angle between the field and the normal to the coil.
a. Torque parallel to the short side of the loop,
The length of the short side is 6cm=0.06m.
This is actually the maximum possible torque, when the field is in the plane of the loop.
τ=NiABsinθ
τ=1×3×0.006×4×10^-3
τ=0.072 ×10^-3
τ=7.2 ×10^-5Nm
b. Torque parallel to the long side of the loop,
The length of the short side is 10cm=0.1m.
This is actually the maximum possible torque, when the field is in the plane of the loop.
τ=NiABsinθ
τ=3×0.006×4×10^-3
τ=0.072 ×10^-3
τ=7.2 ×10^-5Nm.
c. Torque perpendicular to the plane. When the field is perpendicular to the loop the torque is zero.
If the torque is perpendicular to the plane, we said theta is the angle between the normal and the magnetic field
Then if the torque is perpendicular to the normal, then the angle between the torque and the normal is 0, the sinθ = 0
τ=NiABsinθ
Since sinθ =0
Then,
τ=0Nm
a 2.0 kg mass moving to the east at a speed of 4.0 m/s collides head-on in a perfectly inelastic collision with a stationary 2.0 kg mass. how much kinetic energy is lost during
Answer:
12J
Explanation:
Kinetic Energy before collision = 1/2mv1^2 = 1/2×2×4^2 = 16J
Velocity after collision (v2) = m1v1/m1+m2 = 2×4/2+2 = 8/4 = 2m/s
Kinetic Energy after collision = 1/2mv2^2 = 1/2×2×2^2 = 4J
Kinetic Energy lost = 16J - 4J = 12J
Answer:
Lost in kinetic energy = 12 J
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m')..................................... Equation 1
Where m = mass of first body, u = initial of the first body, m' = mass of the second body, u' = initial velocity of the second body, V = common velocity.
Making V the subject of the equation,
V = mu+m'u'/(m+m')........................... Equation 2
Where m = 2.0 kg, m' = 2.0 kg, u = 4.0 m/s, u' = 0 m/s ( stationary).
Substitute into equation 2
V = (2×4 + 2×0)/(2+2)
V = 8/4
V = 2 m/s.
Total kinetic energy before collision = 1/2mu² = 1/2(2)(2)² = 16 J.
Total Kinetic energy after collision = 1/2V²(m+m') = 1/2(2²)(4) = 4 J.
Thus
Lost in kinetic energy = 16 - 4
Lost in kinetic energy = 12 J
Wile E. Coyote is once again pursuing the Roadrunner, chasing the bird in a rocket-powered car. Unsurprisingly, the Roadrunner outsmarts him, and he sails off a cliff in the desert. His velocity when he leaves the cliff is horizontal with a magnitude of 24m/s, but the rocket continues to provide a constant horizontal acceleration of 3m/s2 . The cliff is 29 meters tall. How far from the base of the cliff does the coyote crash into the ground? Assume the ground is flat.
Answer:
The coyote crashes 66 m from the base of the cliff.
Explanation:
Hi there!
The equation of the position vector of the Coyote is the following:
r = (x0 + v0 · t + 1/2 · a · t², y0 + 1/2 · g · t²)
Where:
r = postion vector of the Coyote at time t.
x0 = initial horizontal position.
v0 = initial horizontal velocity.
t = time.
a = horizontal acceleration.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Let's place the origin of the frame of reference at the edge of the cliff so that x0 and y0 = 0.
When the Coyote reaches the ground, the vertical component of its position vector (r1 in the figure) will be equal to -29 m. When the vertical component of the position vector is -29 m, the horizontal component will be equal to the horizontal distance traveled by the Coyote (r1x in the figure). So, let's find the time at which the y-component of the position vector is -29 m:
y = y0 + 1/2 · g · t² (y0 = 0)
-29 m = -1/2 · 9.8 m/s² · t²
t² = -29 m / -4.9 m/s²
t = 2.4 s
Now, let's find the x-component of the vector r1 in the figure:
x = x0 + v0 · t + 1/2 · a · t² (x0 = 0)
x = 24 m/s · 2.4 s + 1/2 · 3 m/s² · (2.4 s)²
x = 66 m
The coyote crashes 66 m from the base of the cliff.
The problem involves determining the distance Wile E. Coyote travels horizontally before he crashes into the ground. It involves the use of motion equations to first calculate the time it takes for the coyote to hit the ground and then the distance it travels horizontally. The calculated distance is roughly 65.4 meters.
Explanation:In this problem, we are asked to calculate the distance that Wile E. Coyote travels along the ground before he crashes.
To solve this, we need to use the laws of motion. First, since the coyote falls with a constant acceleration due to gravity, we can use the equation of motion to determine the time it takes for him to hit the ground:
29 = 0.5 * g * t2
By plugging in the acceleration due to gravity (g=9.8m/s2), we determine that t is approximately 2.44 seconds.
Next, we have to calculate how far the coyote travels horizontally. He starts with a velocity of 24m/s, but he also accelerates due to the rocket. So the distance he travels is given by:
x = v0t + 0.5*a*t2
Substituting the known values (v0 = 24m/s, a = 3m/s2, t = 2.44s), we find that the coyote travels roughly 65.4 meters along the ground before he crashes into it.
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Suppose a negative point charge is placed at x = 0 and an electron is placed at some point P on the positive x-axis. What is the direction of the electric field at point P due to the point charge, and what is the direction of the force experienced by the electron due to that field?
O E along -X; F along –X
O E along +x; F along +x
O E along –x; F along +x
O E along +x; F along - x
Answer:
E along –x; F along +x
Explanation:
When a negative point charge is placed at x=0 and an electron is place at any point P on the positive x-axis the as we know that the like charges repel each other, but there will be no change in the natural tendency of the individual electric field lines. So the direction of the electric field lines at point P due to the point charge will be towards the negative x-axis.The direction of force on the electron due to the electric field of point charge at x=0 will be towards positive x-axis in accordance of the repulsion effect.The electric field at point P due to the negative point charge is directed along the -x axis, and the force experienced by the electron at point P due to this field is also along the -x axis.
Explanation:The electric field at point P due to the negative point charge at x = 0 is directed along the -x axis. This is because electric field lines always point away from positive charges and toward negative charges. Since the negative charge is located at x = 0, the electric field at point P, which is on the positive x-axis, points in the opposite direction, i.e., along the -x axis.
The force experienced by the electron at point P due to this electric field will be in the same direction as the electric field, i.e., along the -x axis. Like charges repel each other, so the negative point charge will exert a repulsive force on the electron.