Answer:
On moon time period will become 2.45 times of the time period on earth
Explanation:
Time period of simple pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex] ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth
As when we go to moon, acceleration due to gravity on moon is [tex]\frac{1}{6}[/tex] times os acceleration due to gravity on earth
So time period of pendulum on moon is equal to
[tex]T_{moon}=2\pi \sqrt{\frac{l}{\frac{g}{6}}}=2\pi \sqrt{\frac{6l}{g}}[/tex] --------eqn 2
Dividing eqn 2 by eqn 1
[tex]\frac{T_{moon}}{T}=\sqrt{\frac{6l}{g}\times \frac{g}{l}}[/tex]
[tex]T_{moon}=\sqrt{6}T=2.45T[/tex]
So on moon time period will become 2.45 times of the time period on earth
Final answer:
The period of a pendulum on the Moon would be longer because the Moon's gravity is weaker. To achieve the same one-second period as on Earth, a pendulum needs to be much shorter due to the Moon's 1/6th gravitational acceleration. Consequently, a pendulum's frequency would decrease if taken from Earth to the Moon.
Explanation:
The period of a simple pendulum is affected by the acceleration due to gravity, which is less on the Moon than on Earth. Hence, if you took a pendulum clock, like a grandfather clock, to the Moon, its pendulum would swing more slowly because of the Moon's weaker gravity. To maintain a steady tick-tock of one second per period on the Moon, the pendulum would need to be much shorter. A grandfather clock pendulum designed to have a two-second period on Earth with a length of 50 cm would need to be only 8.2 cm long on the Moon to achieve the same period, since the Moon's gravity is 1/6th that of Earth. Therefore, if a pendulum from Earth was taken to the Moon, its frequency would decrease because the acceleration due to gravity on the Moon is less than that on Earth.
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0-kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. (a) How high is the balloon when the rock is thrown? (b) How high is the balloon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.
Answer:
a)-296 m/s
b)-176 m
c) 198 m
Explanation:
given data:
[tex]v_{oy}\\[/tex]=-20 m/s
[tex]v_{x}[/tex]= 15 m/s
t = 6 s
To find
a)[tex]H_{1}[/tex]
b)[tex]H_{2}[/tex]
c) d
d) find the velocity of observer at rest in basket and rest on ground
a) according to kinematic equation the displacement is given by:
[tex]H_{1}[/tex]=[tex]v_{oy}t\\[/tex]-[tex]\frac{1}{2} gt^{2}[/tex]..............(1)
putting values in 1
[tex]H_{1}[/tex]=-296 m/s
negative sign is due to direction and also we choose to take off point to be the origin
b) while the stone was travelling [tex]H_{1}[/tex] for 6 s so the ballon was also travelling displacement [tex]H_{b}[/tex] with [tex]v_{oy}\\[/tex]
[tex]H_{b}[/tex] =[tex]v_{oy}t\\[/tex]
=-6*20
=-120 m
the height above the earth is given by
[tex]H_{2}[/tex]=[tex]H_{1}[/tex]-[tex]H_{2}[/tex]= -176 m
c) the horizontal displacement is given by
x=[tex]v_{x}t\\[/tex]=15*6=90 m
so the distance between balloon and stone is
d=[tex]\sqrt{H_{1}^{2}+x^{2} }[/tex]=198 m
d) the velocity component of the balloon :
1) relative to person rest in the balloon are given by
The vertical component:
[tex]v_{yr} =v_{by} +v_{py}[/tex]=[tex](-gt)-0=9.8*6=-58.8 m/s[/tex]
The horizontal component:
[tex]v_{xr} =v_{bx} +v_{px}= 15-0=15m/s[/tex]
2) relative to person rest in the earth are given by
The vertical component:
[tex]v_{yr} =v_{by} -v_{py}=(-gt)-20=9.8*6-20=-78.8 m/s[/tex]
The horizontal component:
[tex]v_{xr} =v_{bx} -v_{px}= 15-0=15m/s[/tex]
A police car in a high-speed chase is traveling north on a two-lane highway at 35.0 m/s. In the southbound lane of the same highway, an SUV is moving at 18.0 m/s. Take the positive x-direction to be toward the north. Find the x-velocity of the police car relative to the SUV.
Answer:
53 m/s
Explanation:
Since we take the positive x-direction to be toward the north, an SUV travelling to the south would have a negative velocity, -18m/s, relative to Earth. Velocity of Earth relative to the SUV would be 18m/s. And velocity of police car relative to Earth would be positive, 35 m/s.
Velocity of police relative to SUV would equal to velocity of police car relative to Earth plus velocity of Earth relative to SUV
= 35 + 18 = 53 m/s
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
If a seed is launched at an angle of 0° with the maximum initial speed, how far from the plant will it land? Ignore air resistance, and assume that the ground is flat. (a) 20 cm; (b) 93 cm; (c) 2.2 m; (d) 4.6 m.
The seed launched at an angle of 0° will land 0 cm from the plant regardless of the initial speed because in a projectile motion, angle of 0° results in zero horizontal distance traveled.
Explanation:This question can be solved by using the formulas of projectile motion. In such motion, the horizontal distance traveled by an object (range) can be calculated using the formula:
R = (v² sin(2θ))/g
where 'v' is the speed of the object, 'θ' is the angle of projection, and 'g' is the acceleration due to gravity.
Here, the launch angle is 0° and the speed is 4.6 m/s. Since the sin(2*0°) is 0, whatever the speed would be, the seed will drop straight down, having no horizontal distance traveled.
So the seed will not move horizontally and it will land 0 cm away from the plant (option not given in choices).
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To solve this problem, we can use the kinematic equations of motion. When a projectile is launched at an angle from the horizontal, we can split its initial velocity into horizontal and vertical components. Since the launch angle is 0°, the initial velocity will have only a horizontal component, and the vertical component will be zero.
Given:
Initial speed,
v 0=4.6m/s
Launch angle,
θ=0∘
Initial height,
h=20cm=0.20m
Acceleration due to gravity,
g= 9.8m/s²
First, we find the time
t it takes for the seed to hit the ground:
Since the initial vertical velocity is 0, we can use the equation:
h= 1/2gt²
Substituting the values:
0.20= 1/2×9.8×t²
0.20=4.9t²
Solving for:
t²= 4.9/0.20
t²=0.040816
t²≈0.202s
Now, we can use this time to find the horizontal distance travelled x. Since the launch angle is 0°, the horizontal velocity remains constant:
x=v₀×t
x=4.6×0.202
x≈0.9292m
So, the seed will land approximately 0.9292m away from the plant.
The closest option provided is (b) 93 cm. However, the correct value is 92.92 cm, which would round to 93 cm. Therefore, the correct answer is (b) 93 cm.
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a charge of-29.0 μC .
A Find the electric field just inside the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward.
B Find the electric field just outside the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward
C Find the electric field 6.00
cm outside the surface of the paint layer.
Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward.
A) The electric field inside the paint layer is zero
B) The electric field just outside the paint layer is [tex]3.2\cdot 10^7 N/C[/tex] (radially inward)
C) The electric field at 6.00 cm from the surface is [tex]1.2\cdot 10^7 N/C[/tex] (radially inward)
Explanation:
A)
We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:
[tex]\int EdS = \frac{q}{\epsilon_0}[/tex]
where
E is the magnitude of the electric field
dS is the element of the surface
q is the charge contained within the surface
[tex]\epsilon_0[/tex] is the vacuum permittivity
By taking a sphere centered in the origin,
[tex]\int E dS = E \cdot 4\pi r^2[/tex]
where [tex]4\pi r^2[/tex] is the surface of the Gaussian sphere of radius r.
In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than
[tex]R=9.0 cm = 0.09 m[/tex] (radius of the plastic sphere is half of the diameter)
Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:
[tex]q=0[/tex]
And therefore,
[tex]E4\pi r^2 = 0\\\rightarrow E = 0[/tex]
So, the electric field inside the plastic sphere is zero.
B)
Here we apply again Gauss Law:
[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]
In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so
[tex]r=R=0.18 m[/tex]
The charge contained within the Gaussian sphere is therefore
[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]
Therefore, the electric field is
[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C[/tex]
And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is
[tex]E=3.2\cdot 10^7 N/C[/tex]
C)
For this part again, we apply Gauss Law:
[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]
In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be
r = 9 cm + 6 cm = 15 cm = 0.15 m
While the charge contained within the sphere is again
[tex]q=-29.0 \mu C = -29.0\cdot 10^{-6}C[/tex]
Therefore, the electric field in this case is
[tex]E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C[/tex]
And again, this is radially inward, so according to the sign convention asked in the problem,
[tex]E=1.2\cdot 10^7 N/C[/tex]
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To find the electric field just inside the paint layer, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium.
A) Inside the paint layer:
Since the paint is spread in a very thin uniform layer over the surface of the plastic sphere, we can consider a Gaussian surface just inside the paint layer, with a radius slightly smaller than the sphere's radius (let's call it r). The charge enclosed within this Gaussian surface is the charge on the sphere, which is -29.0 μC.
Gauss's law equation can be written as:
Φ = Q_enclosed / ε₀
The electric field just inside the paint layer (E₁) is given by:
E₁ * A = Q_enclosed / ε₀
E₁ * 4πr² = -29.0 μC / ε₀
E₁ = (-29.0 μC / ε₀) / (4πr²)
The direction of the electric field just inside the paint layer will be radially outward, opposite to the direction of the charge.
B) Just outside the paint layer:
The electric field just outside the paint layer (E₂) will be the same as the electric field inside the sphere when it was uncharged. This is because the charge on the paint layer is spread on the outer surface of the sphere.
So, E₂ = E₁ (since it's radially outward)
C) 6.00 cm outside the surface of the paint layer:
To find the electric field 6.00 cm outside the surface of the paint layer, we need to calculate the electric field due to the charge on the sphere.
E₃ = k * Q / r²
where k is the Coulomb's constant (k = 9.0 x 10^9 N m²/C²), Q is the charge on the sphere (-29.0 μC), and r is the distance from the center of the sphere to the point where we want to find the electric field (6.00 cm = 0.06 m + radius of the sphere).
Remember that the electric field due to a charged object decreases with the square of the distance.
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A woman exerts a total force of 7 pounds in a horizontal direction on a box as she pushes it up a ramp that is 10 feet long and inclined at an angle of 30 degrees above the horizontal. Find the work done on the box.
Answer:
W = 60.62 ft-lbs
Explanation:
given,
Horizontal force = 7 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
Work done is equal to force into displacement.
W = F.d cos θ
W = 7 x 10 x cos 30°
W = 70 x 0.8660
W = 60.62 ft-lbs
Hence, work done on the box is equal to W = 60.62 ft-lbs
A total charge of 6.3 x 10-8 C is placed on a 2.7 cm radius isolated conducting sphere. The surface charge density is: A. 2.55 x 10-4 C/m3 Explain B. 8.64 x 10-5 C/m2 C. 2.75 x 10-5 C/m2 D. 6.88 x 10-6 C/m2 E. 7.43 x 10-7 C/m
Answer:
[tex]6.88\times 10^{-6}\ C/m^2[/tex]
Explanation:
Q = Total charge = [tex]6.3\times 10^{-8}\ C[/tex]
[tex]\rho[/tex] = Surface charge density
r = Radius = 2.7 cm
A = Surface area = [tex]4\pi r^2[/tex]
Charge is given by
[tex]Q=A\rho\\\Rightarrow Q=4\pi r^2\times \rho\\\Rightarrow \rho=\dfrac{Q}{4\pi r^2}\\\Rightarrow \rho=\dfrac{6.3\times 10^{-8}}{4\pi (2.7\times 10^{-2})^2}\\\Rightarrow \rho=0.00000687706544224\\\Rightarrow \rho=6.88\times 10^{-6}\ C/m^2[/tex]
The surface charge density is [tex]6.88\times 10^{-6}\ C/m^2[/tex]
The period of a carrier wave is T=0.005 seconds. Determine the frequency and wavelength of the carrier wave.
Answer:
f = 200[Hz]; L=1500[km]
Explanation:
We know that frequency is the reciprocal of the period, therefore.
[tex]f=\frac{1}{T} \\f=1/0.005\\f=200[Hz][/tex]
And the wavelength is determined using the following equation.
[tex]L=\frac{c}{f} \\where:\\c = light speed = 3*10^{8}[m/s]\\ f = frecuency [Hz]\\L = wavelength [m]\\L= 3*10^{8}/ 200\\ L = 1500000[m] = 1500[km][/tex]
Final answer:
The frequency of a carrier wave with a period of 0.005 seconds is calculated to be 200 Hz. The wavelength cannot be determined without additional information on the wave's speed.
Explanation:
The question involves calculating the frequency and wavelength of a carrier wave given its period (T=0.005 seconds). The frequency (f) of a wave is the reciprocal of its period, defined by the equation f = 1/T. Therefore, for a period of 0.005 seconds, the frequency would be f = 1/0.005 Hz, which equals 200 Hz. To determine the wavelength (λ), we would need the speed (v) of the wave, which is typically the speed of light (c) for electromagnetic waves, but since the speed and specific type of wave are not provided, we can't calculate the wavelength directly from the given information here.
A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.50 m above. The second student catches the keys 1.10 s later. (a) With what initial velocity were the keys thrown? magnitude m/s direction (b) What was the velocity of the keys just before they were caught? magnitude m/s direction
Answer:
a) 8.58 m/s upward
b) -2.211 m/s downward
Explanation:
Let gravitational acceleration g = -9.81m/s2. This is negative because it deceleration the upward motion of the key.
(a)We have the following equation of motion
[tex]s = v_0t + gt^2/2[/tex]
where [tex]v_0[/tex] is the initial upward velocity of the keys, t = 1.1s is the time it takes for the keys to travel a distance of s = 3.5 m
[tex]3.5 = v_01.1 - 9.81*1.1^2/2[/tex]
[tex]3.5 = 1.1v_0 - 5.94 [/tex]
[tex]1.1v_0 = 3.5 + 5.94 = 9.44[/tex]
[tex]v_0 = 9.44 / 1.1 = 8.58 m/s[/tex]
So the keys were thrown initially upward with a speed of 8.58 m/s
(b) If the initial velocity of the key is 8.58 m/s and it is subjected to a deceleration of 9.81m/s2 for 1.1s then the velocity right at the 1.1s instant is
[tex]v = v_0 + gt = 8.58 - 9.81*1.1 = -2.211 m/s[/tex]
So they keys would have a downward speed of 2.211 m/s
According to a simplified model of a mammalian heart, at each pulse approximately 20 g of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heat muscle?
Answer:
0.02 N.
Explanation:
Given:
m = 20 g
= 0.02 kg.
vi = 0.35 m/s
vo = 0.25 m/s
t = 0.1 s
Force, F = m * a
Where,
m = mass of the blood
a = acceleration.
Acceleration, a = (vi - vo)/t
Where,
vi = final velocity
vo = initial velocity
t = time.
a = (0.35 - 0.25)/0.1
= 0.1/0.1
= 1 m/s².
F = 0.02 * 1
= 0.02 N.
My sling shot shoots pellets at 50m/s. Find two angles of elevation that can be used to hit a target 65m away. (Assume air resistance is negligible. Assume the gravitational constant ????=10m/s2 for this problem.)
Answer:
Explanation:
Given
velocity of launch [tex]u=50\ m/s[/tex]
Target is [tex]R=65\m\ away[/tex]
Suppose [tex]\theta [/tex] is the launch angle
We know Range of Projectile is
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]65=\frac{50^2\times \sin 2\theta }{9.8}[/tex]
[tex]\sin 2\theta =0.254[/tex]
because [tex]\sin \theta =\sin (180-\theta )[/tex]
so either [tex]2\theta =14.71[/tex]
[tex]\theta =7.35^{\circ}[/tex]
or [tex]180-2\theta =14.71[/tex]
[tex]\theta =82.64^{\circ}[/tex]
A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00 km. After dropping off a package, he drives 9.50 km 35.0° north of east to Starbucks. What is the direction relative to the positive x axis?
Answer:
[tex]\theta=7^o[/tex]
Explanation:
Displacement
It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.
The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is
[tex]\vec r_1=<0,2>\ km[/tex]
Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by
[tex]\vec r_2=<7cos60^o,-7sin60^o>\ km=<3.5\ ,-6.06>\ km[/tex]
Finally, he drives 9.5 km 35° north of east.
[tex]\vec r_3=<9.5cos35^o,9.5sin35^o>\ km=<7.78\ ,\ 5.45>\ km[/tex]
The total displacement is
[tex]\vec r_t=<0,2>\ km+<3.5\ ,-6.06>\ km+<7.78\ ,\ 5.45>\ km[/tex]
[tex]\vec r_t=<11.28,1.39>\ km[/tex]
The direction can be calculated with
[tex]\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232[/tex]
[tex]\boxed{\theta=7^o}[/tex]
A physics demo launches a ball horizontally while dropping a second ball vertically at exactly the same time. Trajectories of two balls. The balls are initially placed at the same height above the ground. The first ball moves vertically downward, and hits the ground directly below the initial point. The second ball is launched horizontally, and follows a parabolic trajectory until it hits the ground at some horizontal distance from the initial point. Which ball hits the ground first?
Answer:
The two balls will hit the ground at the same time neglecting air resistance.
Explanation:
The vertical motion of the two ball are independent of each other. Also, the balls are falling from the same height in the same time. The ball projected horizontal neglecting air resistance is traveling with a constant velocity ( the same distance in the equal time) has only force of gravity acting on it. They will therefore hit the ground at the same time because they are acted upon by the same acceleration in the same time from the same height.
The vertical and horizontal motions of an object are independent. Hence, despite different trajectories, a ball dropped vertically and a ball launched horizontally from the same height will hit the ground simultaneously, as exemplified by Galileo's thought experiment.
Explanation:In the realm of Physics, an interesting question has been asked: Which ball hits the ground first - a ball that is dropped vertically or a ball that is launched horizontally? According to the principle of independence of motion, the horizontal motion and the vertical motion of an object are independent of each other. It implies that the horizontal velocity of a body has no effect on its vertical velocity. Thus, the ball dropped directly and the ball launched horizontally would strike the ground at the same time. Both balls are subjected to the same gravitational acceleration (g), without any initial vertical velocity.
This concept is a perfect illustration of Galileo's thought experiment, whereby he proved that the time for the balls to hit the ground is determined by their vertical motion alone and that horizontal motion does not affect the descent time.
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A 5 kg ball approaches a wall at a speed of 4 m/s. It then bounces off the wall in the opposite direction at the same speed. What is the magnitude of the average force exerted on the ball if it is in contact with the wall for 0.1 s?
Answer:
Average force = 2 kg m/s²
Explanation:
Given
mass = 5 kg
initial velocity = 0 m/s
final velocity = 4 m/s
time = 0.1 sec
Find
average force = ?
Formula
Average force = m (final velocity - initial velocity)/t₂- t₁
= (5kg)(0 m/s - 4 m/s)/0 - 0.1
= 5 kg (- 4 m/s)/(- 0.1 sec)
= 2 kg m/s²
Final answer:
The magnitude of the average force exerted on the ball by the wall is 400 N, computed using the impulse-momentum theorem and the ball's change in momentum during the collision.
Explanation:
To calculate the magnitude of the average force exerted on the ball, we need to first determine the change in momentum of the ball and then use the impulse-momentum theorem. The ball changes direction after hitting the wall, so the final velocity is -4 m/s (negative sign indicating the opposite direction) and the initial velocity is 4 m/s. The total change in velocity is thus -4 m/s - 4 m/s = -8 m/s. The change in momentum (impulse) is given by mass × change in velocity, which is 5 kg × -8 m/s = -40 kg·m/s. The impulse experienced by the ball equals the average force exerted on the ball multiplied by the time of contact, so:
Average Force = - Impulse / Time
Therefore, the magnitude of the average force is:
|Average Force| = 40 kg·m/s / 0.1 s = 400 N.
The negative sign indicates that the force is in the opposite direction of the initial motion, but since the question asks for the magnitude, we take the absolute value.
Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field of 46 N/C. The density of the oil is 0.085 g/cm3 . Calculate for Millikan the charge on the droplet. Is quantization of charge obeyed within the precision of this experiment?
Answer:
The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].
Yes, quantization of charge is obeyed within experimental error.
Explanation:
Given that,
Radius = 1.6μm
Electric field = 46 N/C
Density of oil = 0.085 g/cm³
We need to calculate the charge on the droplet
Using formula of force
[tex]F= qE[/tex]
[tex]mg=qE[/tex]
[tex]V\times\rho\times g=qE[/tex]
[tex]q=\dfrac{V\times\rho}{E}[/tex]
[tex]q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}[/tex]
Put the value into the formula
[tex]q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}[/tex]
[tex]q=3.106\times10^{-16}\ C[/tex]
We need to calculate the quantization of charge
Using formula of quantization
[tex]n = \dfrac{q}{e}[/tex]
Put the value into the formula
[tex]n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}[/tex]
[tex]n=1941.25[/tex]
Yes, quantization of charge is obeyed within experimental error.
Hence, The charge on the droplet is [tex]3.106\times10^{-16}\ C[/tex].
Yes, quantization of charge is obeyed within experimental error.
Answer:
3.11 x 10^-16 C
Explanation:
radius of drop, r = 1.6 micro metre = 1.6 x 10^-6 m
Electric field, E = 46 N/C
density of oil, d = 0.085 g/cm³ = 85 kg/m³
Let the charge on the oil drop is q.
So, the weight of the drop is balanced by the electric force on the drop.
mass of drop x g = charge of the drop x electric field
m x g = q E
Volume x density x g = q E
[tex]\frac{4}{3} \pi\times r^{3}\times d\times g = q \times E[/tex]
[tex]\frac{4}{3} \pi\times 1.6^{3}\times 10^{-18}\times 85\times 9.8 = q \times 46[/tex]
q = 3.11 x 10^-16 C
Thus, the charge on the oil drop is 3.11 x 10^-16 C.
If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?
Answer:
+5.4×10⁻⁷ C
Explanation:
Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)
The formula for potential is
V = kq/r............................ Equation 1
Where V = electric potential, k = proportionality constant, q = charge, r = distance.
making q the subject of the equation,
q = Vr/k............................ Equation 2
Given: V = 490 V, r = 10 m,
Constant: k = 9×10⁹ Nm²/C²
Substitute into equation 2
q = 490(10)/(9×10⁹)
q = 5.4×10⁻⁷ C
q = +5.4×10⁻⁷ C
Hence the charge is +5.4×10⁻⁷ C
Charge 9 × 10−18 C is on the y axis a distance 5 m from the origin and charge 9 × 10−18 C is on the x axis a distance d from the origin. What is the value of d for which the x component of the force on 9 × 10−18 C is the greatest? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 .
Answer:
d = 3.53 m
Explanation:
The Coulomb Force is given as
[tex]\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r[/tex]
The x-component of the force is equal to
[tex]F_x = F\cos(\theta) = F\frac{x}{\sqrt{x^2 + y^2}} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{(5^2 + d^2)}\frac{d}{\sqrt{5^2 + d^2}} = \frac{1}{4\pi\epsilon_0}\frac{dq_1q_2}{(5^2 + d^2)^{3/2}}[/tex]
This is basically a function of (d). So, the maximum value of this function is the point where its derivative with respect to d is equal to zero.
[tex]\frac{dF_x}{dd} = \frac{kq_1q_2}{(d^2 + 5^2)^{3/2}} - \frac{3d^2kq_1q_2}{(d^2 + 5^2)^{5/2}} = 0\\3d^2 = d^2 + 5^2\\2d^2 = 25\\d = 3.53~m[/tex]
The value of d for which the x component of the force on the charge is the greatest, in accordance to Coulomb's Law, is when d equals 5 meters. At this distance, the x and y components of the force are equal, thus maximizing the x component.
Explanation:The scenario you described involves the concept of Coulomb's Law which states the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Looking at your question, the value of d for which the x component of the force on 9×10−18 C is the greatest would be when d equals 5 meters. The charges would then form an equilateral triangle with the origin, meaning the x component and y component of the force would be equal, hence maximizing the x component.
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Acetone, a component of some types of fingernail polish, has a boiling point of 56°C. What is its boiling point in units of kelvin? Report your answer to the correct number of significant figures.
Answer:
The boiling point of Acetone is 329K (in 3 significant figures)
Explanation:
Boiling point of Acetone = 56°C = 56 + 273K = 329K (in 3 significant figures)
Answer: using the formula 0°C + 273.15 = 273.15K the boiling point in units of kelvin to significant figures is 329.15k.
Explanation: The boiling point of a substance ( acetone) is the temperature at which the vapour pressure of the liquid substance equals the pressure surrounding it. The boiling point of acetone serves as it's characteristic physical properties. This is measured in degree Celsius (°C ) which can be converted to units of Fahrenheit or kelvin. To convert degree Celsius to kelvin this formula is used: 0°C + 273.15 = 273.15K . Given that acetone has boiling point of 56°C,from the formula 0°C is substituted for 56°C. This gives us:
56°C + 273.15= 319.15k.
Also,measurements given in Kelvin will always be larger numbers than in Celsius and the Kelvin temperature scale does not use the degree (°) symbol because Kelvin is an absolute scale, based on absolute zero, while the zero on the Celsius scale is based on the properties of water. I hope this helps. Thanks
What key ingredient in the modern condensation theory was missing or unknown in the nebula theory?
Answer:
Interstellar dust
Explanation:
In modern solar system theory of condensation, interstellar dust, which was lacking in nebular theory, would be the essential component.Cosmic dust is dust that exists in outer space or that has fallen to earth, also called extra-terrestrial dust or spatial powder. The majority of the cosmic dust particle sizes range from a few molecules to 0.1 μm.
Which of the cities that we examined experienced the least significant temperature increase (smallest slopes)?A. Seattle, WA B. Springfield, IL C. Phoenix, AZ D. New York, NY
Springfield has the smallest slope.
Answer: Option B.
Explanation:
Springfield is a city in the United States of America. This city is in the states of Massachusetts. It is besides the river Connecticut river which adds to the beauty of this state.
Springfield is the nick name of the city "The queen of the Ozarks". The other name of this city is "the cultural center of the Ozarks". It has a beautiful scenery and this scenic beauty adds to the tourist attraction to a lot of people all around the world.
Particle physicists use particle track detectors to determine the lifetimeof short-lived particles. A muon has a mean lifetime of 2.2sand makes a track 9.5 cm long before decayinginto an electron and two neutrinos. What was the speed of the muon?
Final answer:
To calculate the speed of a muon before it decays, the distance the muon travels and the time it takes are used in the formula L = v x t. For a distance of 9.5 cm and a time of 2.20 microseconds, the speed is found to be approximately 43,182 meters per second.
Explanation:
The question is seeking to calculate the speed of a muon before it decays. Assuming the mean lifetime of a muon is 2.20 microseconds (us) and the muon makes a track 9.5 cm long in this time, we can calculate the muon's speed. Let's consider the formula:
L = v x t
where L is the distance traveled, v is the velocity, and t is the time. Given L = 9.5 cm (or 0.095 m) and t = 2.20 x 10-6 s, we plug in the values:
v = L/t
v = 0.095 m / 2.20 x 10-6 s
v = 4.31818 x 104 m/s
The muon's speed was approximately 43,182 m/s.
A cathode-ray tube (CRT) is an evacuated glass tube. Electrons are produced at one end, usually by the heating of a metal. After being focused electromagnetically into a beam, they are accelerated through a potential difference, called the accelerating potential. The electrons then strike a coated screen, where they transfer their energy to the coating through collisions, causing it to glow. CRTs are found in oscilloscopes and computer monitors, as well as in earlier versions of television screens.If the accelerating potential is 95.0 V, how fast will the electrons be moving when they hit the screen?
Answer: v = 5.79 * 10^10m/s.
Explanation: By using the work-energy theorem, we know that the work done on the electron by the potential difference equals the kinetic energy of the electrons.
Mathematically, we have that
qV = 1/2mv²
q= magnitude of an electronic charge = 1.609*10^-16c
V= potential difference = 95v
m = mass of an electronic charge = 9.11* 10^-31kg.
v = velocity of electron.
Let us substitute the parameters, we have that
1.609*10^-16 * 95 = (9.11*10^-31 * v²) /2
1.609*10^-16 * 95 * 2 = 9.11*10^-31 * v²
305.71 * 10^-16 = 9.11 * 10^-31 * v²
v² = 305.71 * 10^-16/ 9.11 * 10^-31
v² = 3.355 * 10^21.
v = √3.355 * 10^21
v = 5.79 * 10^10 m/s
Final answer:
Using the principle of energy conservation and the equation KE = qV = ½ mv^2, the velocity of the electrons in a CRT with an accelerating potential of 95.0 V is calculated to be approximately 5.8 x 10^6 m/s.
Explanation:
To determine how fast electrons will be moving when they hit the screen in a Cathode-Ray Tube (CRT) with an accelerating potential of 95.0 V, one can use the principle of energy conservation. The amount of kinetic energy (KE) gained by an electron when it is accelerated through a potential difference (V) is equal to the change in electrical potential energy (PE), which is given by the equation KE = qV, where q is the charge of the electron (1.602 x 10-19 Coulombs) and V is the potential difference.
The formula for kinetic energy is KE = ½ mv^2, with 'm' representing the mass of the electron (9.109 x 10^-31 kg) and 'v' being the velocity of the electron. Setting these equations equal gives us qV = ½ mv^2. Solving for 'v' and plugging in the values for q, V, and m, we can find the velocity of the electrons when they hit the screen.
Calculation:
qV = ½ mv^2
2qV = mv^2
v = √(2qV/m)
v = √(2 * 1.602 x 10^-19 C * 95.0 V / 9.10^9 x 10^-31 kg)
v ≈ 5.8 x 10^6 m/s
Thus, the electrons would be traveling at approximately 5.8 x 10^6 m/s when they hit the screen.
If a mass of 0.5 kg is displaced by 30 cm using a spring with a spring constant of 2 N/m, find the following:a) The angular frequencyb) The period of oscillationc) The position at t = 2 secondsd) The velocity at t = 2 secondse) The acceleration at t = 2 seconds
Answer
given,
mass of the block, m = 0.5 Kg
displacement, x = 30 cm = 0.3 m
Spring constant, k = 2 N/m
a) Angular frequency
[tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega = \sqrt{\dfrac{2}{0.5}}[/tex]
[tex]\omega = 2\ rad/s[/tex]
b) Period of oscillation
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\pi}{2}[/tex]
T = 3.14 s
c) Position at t = 2
x = -A cos ω t
A = 0.3 ω = 2 rad/s
x = -0.3 cos (2 x 2)
x = 0.196 m
d) velocity at t= 2
v = A ω sin ω t
v = 0.3 x 2 x sin 4
v = -0.454 m/s
e) acceleration at t= 2
a = A ω² cos ω t
a = 0.3 x 2² cos 4
a = -0.784 m/s²
Two siblings are in a funny mood and push each other away with a force of 20 N. If sibling 1 has a mass of 50 kg and sibling 2 a mass of 60 kg, what will be the ratio between their respective accelerations (sibling 1 : sibling 2)? (2 significant figures)
Answer:
[tex]\large\boxed{\large\boxed{\frac{a_1}{a_2}=1.2}}[/tex]
Explanation:
Newton's second law states the the net force exerted over a body, [tex]F[/tex], is proportional to the product of its mass, [tex]m[/tex] , and its acceleration, [tex]a[/tex] :
[tex]F=m\times a[/tex]
As a consequence of Newton's third law, action and reaction forces are equal in magnitude and opposite in direction. Hence, 20 N is the same magnitude of the forces over each sibling.
1. Sibling 1:
[tex]20N=50kg\times a_1[/tex]
2. Sibling 2:
[tex]20N=60kg\times a_2[/tex]
3. Ratio sibling 1: sibling 2:
[tex]\frac{20N}{20N} =\frac{50kg\times a_1}{60kg\times a_2}[/tex]
[tex]\frac{a_1}{a_2}=\frac{60}{50}=1.2[/tex]
The electric field near the surface of Earth points downward and has a magnitude of 152 N/C. What is the ratio of the magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron?
Answer:
[tex]2.7\times 10^{12}[/tex]
Explanation:
We are given that
Electric field =[tex]E=152N/C[/tex]
We have to find the ratio of magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron.
We know that
F=qE
Charge on an electron,q=[tex]1.6\times 10^{-19}C[/tex]
Using the formula
Upward electric force=[tex]152\times 1.6\times 10^{-19}[/tex] N
Upward electric force=[tex]F_e=2.43\times 10^{-17}[/tex] N
Mass of electron=[tex]m_e=9.1\times 10^{-31} kg[/tex]
[tex]g=9.8m/s^2[/tex]
Gravitational force=[tex]F=mg[/tex]
Using the formula
Gravitational force=[tex]F_g=9.1\times 10^{-31}\times 9.8=8.92\times 10^{-30} N[/tex]
Ratio of Fe to the Fg=[tex]\frac{2.43\times 10^{-17}}{8.92\times 10^{-30}}[/tex]
[tex]\frac{F_e}{F_g}=2.7\times 10^{12}[/tex]
To how many significant figures should each answer be rounded?
Equation:
A: ( 6.626 x 10^− 34 J * s ) ( 2.9979 x 10^8 m / s ) / 4.630 x 10^− 7 m = 4.290299222462 x 10^− 19 J ( unrounded )
1. After rounding, the answer to equation A should have __________.
Answer: After rounding, the answer to equation A should have [tex]4.290\times 10^{-19}J[/tex]
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s preceding the first integers are never significant.
All zero’s after the decimal point are always significant.
The rule apply for the multiplication and division is :
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
Thus for [tex]\frac{6.626\times 10^{-34}Js}{4.630\times 10^{-7}m}\times 2.9979\times 10^8m/s}=4.290299222462\times 10^{-19}J[/tex]
In this problem, 6.626 has 4 significant figures, 4.630 has 4 significant figures and 2.9979 has 5 significant digits thus product will have the least number of significant figures which is 4. So, the answer will be in 4 significant figures. Thus the answer is [tex]4.290\times 10^{-19}J[/tex]
The answer to equation A should have four significant figures, as the least precise measurement in the equation has four significant figures.
Explanation:When you are calculating with measured values, the rule is that your result cannot be more precise than the least precise measurement. This means when using values in calculations, you need to look at the number of significant figures in the original measurements to determine how many significant figures to use in the final answer.
In the given equation, A: (6.626 x 10−34 J * s) (2.9979 x 108 m / s) / 4.630 x 10−7 m = 4.290299222462 x 10−19 J (unrounded), the least precise measurement is 4.630 x 10−7 m, which has four significant figures. Therefore, after rounding, the answer to equation A should also have four significant figures.
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You throw a rock upward. The rock is moving upward. but it is slowing down. If we define the ground as the origin, theposition of the rock is ____ and the velocity of the rock is ____ .A. positive, positiveB. positive, negativeC. negative, positiveD. negative. negative
Answer:
.A. positive, positive.
Explanation:
When we throw a rock upward , it will decelerate due to gravitation . It will have acceleration in downward direction or - ve acceleration in upward direction.
If we define the ground as the origin and upward direction as positive , anything in upward direction will be positive and in downward direction will be negative . For example , in the case described above , acceleration of rock thrown upward is negative because is in downward direction .
Its position is in upward direction , so its position is positive with respect to ground.
It is going in upward direction . So velocity too will be positive.
A potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 1.35 × 10 − 20 J 1.35×10−20 J of work is required to eject a positive sodium ion (Na + ) (Na+) from the interior of the cell, what is the magnitude of the potential difference between the inner and outer surfaces of the cell?
Answer:
Explanation:
Given
Work required [tex]W=1.35\times 10^{-20}\ J[/tex]
Work done to eject the sodium ion from interior of the cell is given by the product of charge and Potential difference between inner and outer surface of the cell.
[tex]W=q\times V[/tex]
Charge on sodium ion [tex]q=1.6\times 10^{19}\ C[/tex]
[tex]V=\frac{W}{q}[/tex]
[tex]V=\frac{1.35\times 10^{-20}}{1.6\times 10^{19}}[/tex]
[tex]V=0.0718\ V[/tex]
The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.
Potential difference:When potential difference is V and charge of ion is q. Then, work required to eject a postive charge from the interior of cell is computed as,
Work = q * V
It is given that, work W = [tex]1.35*10^{-20}J[/tex]
Charge on sodium ion , [tex]q=1.6*10^{-19}C[/tex]
Substituting above values in above relation.
[tex]V=\frac{W}{q}\\ \\V=\frac{1.35*10^{-20} }{1.6*10^{-19} }\\ \\V=0.0844Volt[/tex]
Hence, the magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0844 volt.
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How and why does a day measured with respect to the Sun differ from a day measured with respect to the stars?
Answer:
A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.
Explanation:
A day measured with respect to the sun differ from a day measured with respect to star by 4 minutes. This is because, it takes 4 minutes for the Earth to rotate the extra amount required for the Sun to return to the same place in the sky.
Each day that goes by, the Earth needs to turn a little further for the sun to return to the same place in the sky. And that extra time is about 4 minutes or 240 seconds.
Final answer:
A solar day is longer than a sidereal day by about 4 minutes, due to Earth's orbit around the Sun requiring additional rotation to align the Sun to the same position in the sky. The sidereal day, on the other hand, represents the true rotation period of Earth as it measures the time it takes for a distant star to reappear in the same position in the sky.
Explanation:
A day measured with respect to the Sun differs from a day measured with respect to the stars primarily in its length due to Earth's simultaneous rotation on its axis and revolution around the Sun. The solar day is the period during which Earth rotates on its axis so that the Sun appears at the same position in the sky on consecutive days. It is about 4 minutes longer than a sidereal day, which is the time it takes for a distant star to return to the same position in the sky.
This difference occurs because as Earth rotates, it also moves along its orbital path around the Sun, so it has to rotate a little more for the Sun to be in the same position overhead. The sidereal day is actually the true rotation period of Earth, being 235.9 seconds shorter than 24 hours, while the solar day includes an additional 4 minutes to account for Earth's movement in orbit.
A rhinoceros is at the origin of coordinates at time t1 = 0. For the time interval from t1 = 0 to t2 = 12.0 s, the rhino’s average velocity has x-component - 3.8 m/s and y-component 4.9 m/s. At time t2 = 12.0 s, (a) what are the x- and y-coordinates of the rhino? (b) How far is the rhino from the origin?
Answer:
a) [tex](x_2,y_2)=(-45.6,58.8)[/tex]
b) [tex]s=74.4097\ m[/tex]
Explanation:
Given:
x-component of avg. velocity, [tex]v_x=-3.8\ m.s^{-1}[/tex]x-component of avg. velocity, [tex]v_y=4.9\ m.s^{-1}[/tex]initial position of rhino, [tex](x_1,y_1)=(0,0)[/tex]initial time, [tex]t_1=0\ s[/tex]final time, [tex]t_2=12\ s[/tex]a)
As we know that average velocity is total displacement per unit time.
Position of x-coordinate of rhino:
[tex]v_x=\frac{x_2}{t_2}[/tex]
[tex]-3.8=\frac{x_2}{12}[/tex]
[tex]x_2=-45.6\ m[/tex]
Position of y-coordinate of rhino:
[tex]v_y=\frac{y_2}{t_2}[/tex]
[tex]4.9=\frac{y_2}{12}[/tex]
[tex]y_2=58.8\ m[/tex]
b)
Now the distance from the origin:
[tex]s=\sqrt{x_2^2+y_2^2}[/tex]
[tex]s=\sqrt{(-45.6)^2+(58.8)^2}[/tex]
[tex]s=74.4097\ m[/tex]
The question concerns vector kinematics in physics. The rhinoceros's coordinates at 12 seconds are (-45.6,58.8), and its distance from the origin is approximately 74.2 meters.
Explanation:The phenomenon described in your question essentially relates to Vector Kinematics. The rhinoceros is said to have a velocity in the x-direction of -3.8 m/s and in the y-direction of 4.9 m/s. These velocities are constant and last for 12 seconds.
To find the x- and y-coordinates, we can simply multiply the component velocities by time.
For the x-coordinate:
x = velocity_x * time = -3.8 m/s * 12 s = -45.6 m
And for the y-coordinate:
y = velocity_y * time = 4.9 m/s * 12 s = 58.8 m
So, for part (a) of your question, the rhino's coordinates at t2 = 12.0 s are (-45.6, 58.8).
Now, for part (b), to calculate the rhino's distance from the origin, we can use the Pythagorean theorem which is rooted in the geometrical interpretation of vectors. The distance from the origin (r) can be given by:
r = sqrt[x² + y²] = sqrt[(-45.6 m)² + (58.8 m)²] = approximately 74.2 m
Therefore, the rhino is approximately 74.2 m from the origin.
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A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.00 mm to the bottom of the incline is 3.80 m/s.What is the speed of the block when it is 3.00 m from the top of the incline?
Answer:
Explanation:
Given
Speed of block at bottom is [tex]v=3.8\ m/s[/tex]
Distance traveled [tex]s=6\ m[/tex]
initial velocity is zero
using equation of motion
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](3.8)^2-0=2\times a\times 6[/tex]
[tex]a=1.203\ m/s^2[/tex]
when it is 3 m from top then
[tex]v^2-u^2=2as[/tex]
[tex]v^2-0=2\times 1.203\times 3[/tex]
[tex]v=2.68\ m/s[/tex]
The problem is a physics question on kinematics, which requires calculating the speed of a block half-way down a frictionless incline using the kinematic equation for constant acceleration.
Explanation:The student's question involves finding the speed of a block at a certain point as it slides down a frictionless incline. The kinematics of motion on an inclined plane is the focus here. We are given the speed of the block after traveling 6.00 mm and we need to calculate its speed after it has traveled 3.00 m down the incline.
To solve this, we can use the kinematic equation for constant acceleration, which is stated as:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the block starts from rest, u = 0. We can find the acceleration by using the given final speed and distance from the first part of the trip (6.00 mm to the bottom). We substitute the acceleration into the kinematic equation again using s as 3.00 m to find the speed at that point.