Final answer:
The object is actually located 1.5 m from the van, and it is 2.1 m tall when considering the given focal length and image properties.
Explanation:
- Focal length of the lens (f) = -0.300 m (negative sign indicates a diverging lens)
- Image distance = 0.250 m (positive because the image appears behind the lens)
- Height of the image = 0.350 m
We need to find:
(a) Actual distance (d) of the object from the van.
(b) Actual height (h) of the object.
(a) To find the actual distance of the object from the van (d), we use the thin lens equation:
1/f = 1/d + 1/image distance
Plugging in the given values:
1/-0.300 = 1/d + 1/0.250
Solving for d:
-3.33 = 1/d + 4
-3.33 - 4 = 1/d
-7.33 = 1/d
d = 1/-7.33
d ≈ -0.136 m
The negative sign indicates that the object is located behind the lens.
(b) To find the actual height of the object (h), we use the magnification formula:
m = height of the image / height of the object = -(image distance / object distance)
Given m = -(image distance / object distance), and we have m = height of the image / height of the object, we can write:
-(image distance / object distance) = height of the image / height of the object
Plugging in the given values:
-(0.250 / -0.136) = 0.350 / h
Solving for h:
h = (0.350 * -0.136) / 0.250
h ≈ -0.190 m
The negative sign indicates that the height of the object is inverted.
So, the answers are:
(a) The object is actually located approximately 0.136 m behind the van.
(b) The actual height of the object is approximately 0.190 m.
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.
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 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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Four point charges each having charge Q are located at the corners of a square having sides of length a. (a) Find an expression for the total electric potential at the center of the square due to the four charges. (Use any variable or symbol stated above along with the following as necessary: ke.)
Answer:
[tex]\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Explanation:
Electric Potential of Point Charges
The electric potential from a point charge Q at a distance r from the charge is
[tex]\displaystyle V=k\frac{Q}{r}[/tex]
Where k is the Coulomb's constant. The total electric potential for a system of point charges is equal to the scalar sum of their individual potentials. The potential is not a vector, so there is no direction or vectors to deal with.
We are required to compute the total electric potential in the center of the square. We need to know the distance from each corner to the center. The diagonal of the square is
[tex]d=\sqrt2 a[/tex]
where a is the length of the side.
The distance from any corner to the center is half that diagonal, thus
[tex]\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}[/tex]
The total potential in the center is
[tex]V_t=V_1+V_2+V_3+V_4[/tex]
Please note all the potentials must be calculated including the sign of the charges. Since all the charges are equal to Q, and the distances are the same, the total potential is 4 times the individual potential of each charge.
[tex]V_t=4\times V[/tex]
[tex]\displaystyle V=9\times 10^9 \frac{Q}{\frac{a}{\sqrt{2}}}[/tex]
Operating
[tex]\displaystyle V=9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Thus:
[tex]\displaystyle V_t=4\times 9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
[tex]\boxed{\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}}[/tex]
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 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.
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.
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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NASA communicates with the Space Shuttle and International Space Station using Ku-band microwave radio. Suppose NASA transmits a microwave signal to the International; Space Station using radio waves with a frequency of 18.0 GHz. What is the wavelength of these radio waves?
Answer:
0.0167 m
Explanation:
f = Frequency = 18 GHz
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
When the speed of the wave is divided by the freqeuncy we get the wavelength
Wavelength is given by
[tex]\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{18\times 10^9}\\\Rightarrow \lambda=0.0167\ m[/tex]
The wavelength of the radio waves is 0.0167 m
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 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
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
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.
Describe the motion of the ball as I see it. Is the path taken by the ball different than if I was sitting at rest at home and I threw a ball straight up into the air?
Answer:
Explanation:
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .
As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.
If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.
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]
A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance x away from the gun. The pellet hits the target a vertical distance y below the gun.
(a) Show that the position of the pellet when traveling through the air is given by y = Ax^2, where A is a constant.
(b) Express the constant A in terms of the initial velocity v and the free-fall acceleration g.
(c) If x = 3 m and y = 0.21 m, what is the initial speed of the pellet?
Answer:
a) y = A x² , b) A = - ½ g / v₀², c) v₀ = 15.46 m / s
Explanation:
For this problem of two-dimensional kinematics, we will use that the time to reach the wall is the same
X axis
x = v₀ₓ t
t = x / v₀ₓ
Y Axis
y = [tex]v_{oy}[/tex] t - ½ g t²
As it shoots horizontally the vertical speed is zero
y = - ½ g t²
We replace
y = - ½ g (x / v₀ₓ)²
The initial speed is all horizontal
v₀ₓ = v₀
y = - ½ g / v₀² x²
y = A x²
b) the expression for the constant is
A = - ½ g / v₀²
c) we look for the initial speed
v₀² = - ½ g x² / y
As the object falls below the exit point its height is negative
v₀ = √ (- ½ 9.8 3²/ (-0.21))
v₀ = 15.46 m / s
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.
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
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
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]
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.
A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stays red for 15s, and she wants to reach the light just as it turns green again. It takes her 1.0s to step on the brakes and begin slowing. What is her speed as she reaches the light at the instant it turns green?
Answer:
5.71428571422 m/s
Explanation:
u = Initial velocity = 20 m/s
v = Final velocity
s = Displacement
a = Acceleration
Time taken = 15-1 = 14 s
Distance traveled in 1 second = [tex]20\times 1=20\ m[/tex]
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s[/tex]
The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s
The final speed of the motorist as she reaches the light is 5.72 m/s
To calculate the speed of the motorist as she reaches the light, we need to first find the deceleration of the motorist.
Formula:
S₁ = ut₁.............. Equation 1Where:
S₁ = displacement of the motorist as its slows downu = initial velocityt₁ = time it takes to slow down.Given:
u = 20 m/st₁ = 1Therefore,
S₁ = 20(1) = 20 mThen,
S = 200-20S = 180 m
Where:
S = distance covered by the motorist before the brake is applied.
we use the formula below to calculate the deceleration of the motorist.
S = ut+at²/2........... Equation 2Where:
a = deceleration of the motorist.Given:
u = 20 m/st = 15-1 = 14 sS = 180 mSubstitute these values into equation 2
180 = 20(14)+a(14²)/2180 = 280+98aSolve for a
98a = 180-28098a = -100a = -100/98a = -1.02 m/s²Finally, we use the formula below to get the speed of the motorist as she reaches the light.
v = u+at............ Equation 3Given:
u = 20 m/sa = -1.02 m/s²t = 14 sSubstitute these values into equation 3
v = 20+(-1.02×14)v = 20-14.28v = 5.72 m/s. Hence, The final speed of the motorist as she reaches the light is 5.72 m/s
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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]
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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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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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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As Halley’s comet orbits the sun, its distance from the sun changes dramatically. If the comet’s speed at a distance of 9.8 × 1010 m from the sun is 5.3 × 104 m/s and angular momentum is conserved, what is its speed when it is 3.6 × 1012 m from the sun? Assume the comet can be treated as a point mass. Ignore radial components of momentum
Answer:
The speed of comet is 1442.77 m/s.
Explanation:
Given that,
Speed of comet's [tex]v_{c}= 5.3\times10^{4}\ m/s[/tex]
Distance from the sun [tex]r_{d}=9.8\times10^{10}\ m[/tex]
Distance [tex]r_{far}=3.6\times10^{12}\ m[/tex]
We need to calculate the speed of comet
Using conservation of angular momentum
[tex]L_{f}=L_{i}[/tex]
[tex]I\omega=I\omega[/tex]
Here. [tex]v = r\omega[/tex]
[tex]\omega=\dfrac{v}{r}[/tex]
[tex]mr_{far}^2\times\dfrac{v_{far}}{r_{far}}=mr_{d}^2\times\dfrac{v_{d}}{r_{d}}[/tex]
[tex]r_{far}\times v_{far}=r_{d}\times v_{d}[/tex]
Put the value into the formula
[tex]3.6\times10^{12}\times v_{far}=9.8\times10^{10}\times5.3\times10^{4}[/tex]
[tex]v_{far}=\dfrac{9.8\times10^{10}\times5.3\times10^{4}}{3.6\times10^{12}}[/tex]
[tex]v_{far}=1442.77\ m/s[/tex]
Hence, The speed of comet is 1442.77 m/s.
Final answer:
The speed of Halley's comet when it is 3.6 x 10^12 m away from the sun can be found using the conservation of angular momentum, resulting in a speed of 1.44 x 10^3 m/s.
Explanation:
To find the speed of Halley's comet at a distance of 3.6 × 1012 m from the sun, we will use the conservation of angular momentum. The orbital momentum of an object around the sun is given by L = mrv, where m is the mass of the comet, r is the radius of the orbit (distance from the sun), and v is the velocity of the comet along that radius.
We are given the comet's speed at a distance of 9.8 × 1010 m is 5.3 × 104 m/s. When the comet is at a distance of 3.6 × 1012 m from the sun, we want to find its new speed, v'. Because angular momentum is conserved, the initial and final angular momentum should be equal:
L initial = L final
m r v = m r' v'
Since the mass of the comet, m, cancels out, we are left with:
r v = r' v'
Plugging in the known values, we get:
9.8 × 1010 m × 5.3 × 104 m/s = 3.6 × 1012 m × v'
Therefore:
v' = (9.8 × 1010 × 5.3 × 104) / (3.6 × 1012)
v' = 1.44 × 103 m/s
The speed of Halley's comet at a distance of 3.6 × 1012 m from the sun is thus 1.44 × 103 m/s.
Design an op-amp circuit to provide an output vO =−[2v1 + (v2/2)]. Choose relatively low values of resistors but ones for which the input current (from each input signal source) does not exceed 50 μA for 1-V input signals.
Answer:
Explanation: see attachment
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.
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.
An astronaut is in space with a baseball and a bowling ball. The astronaut gives both objects an equal push in the same direction. Does the baseball have the same inertia as the bowling ball? Why? Does the baseball have the same acceleration as the bowling ball from the push? Why? If both balls are traveling at the same speed, does the baseball have the same momentum as the bowling ball?
Answer:
Explanation:
Given
Astronaut in space gives an equal push to baseball and bowling ball.
Since the mass of the bowling ball is more than the baseball so inertia associated with a bowling ball is more as compared to baseball
Force applied on baseball and bowling ball is the same so the acceleration of baseball will be more because the mass of baseball is less.
[tex]Force=mass\times acceleration[/tex]
If both are traveling with the same speed then momentum associated with them is given by-product of mass and velocity
Since the mass of the bowling ball is more, therefore, the momentum of bowling ball is more as compared to baseball