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hram777 [196]
2 years ago
9

You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill

inclined 17 ∘∘ above the horizontal. The system is the car and Earth. Part A Determine the change in the total energy of the system when the car moves from the bottom to the top. Ignore friction.
Physics
1 answer:
Stella [2.4K]2 years ago
4 0

Answer:

4.325\times10^6J

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of \theta=17^o

Therefore vertical elevation of the car = 260sin\theta=260sin17^o=76.0166m

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = \frac{1}{2} mv^2). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = 5800\times9.81\times76.0166J=4.325\times10^6J

where, g = acceleration due to gravity

h = height/vertical elevation

You might be interested in
which has a greater kinetic energy, a bowling ball that has a mass of 5kg travelling at 6m/s, or a ship that has a mass of 12000
AVprozaik [17]

Answer:bowling ball has greater kinetic energy

Explanation:

Kinetic energy of bowling ball:

mass=m=5kg

Velocity=v=6m/s

Kinetic energy =ke

Ke=0.5 x m x v x v

Ke=0.5 x 5 x 6 x 6

Ke=90J

Kinetic energy of ship:

mass=m=120000kg

velocity=v=0.02m/s

Ke=0.5 x m x v x v

Ke=0.5 x 120000 x 0.02 x 0.02

Ke=24J

6 0
2 years ago
Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman
Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0
2 years ago
A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca
xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance =  4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

8 0
2 years ago
Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp
erma4kov [3.2K]

Answer:

A = -0.576 μC

B = 4.256 μC

Explanation:

Suppose a single electron charge is 1.6\times10^{-19}C. Then the total charge that is flowing from B to A is:

1.6\times10^{-19} * 1.51 \times 10^{13} = 2.416\times10^{-6}C = 2.416 \mu C

Let A and B be the initial charge of spheres A and B, respectively. Since the net charge is 3.68μC we have the following equation

A + B = 3.68 (1)

When they touch 2.416μC flows from B to A, then they are equal, so we have the following equation

A + 2.416 = B - 2.416

-A + B = 2.416 + 2.416 = 4.832 (2)

Add equation (1) to equation (2) we have

2B = 3.68 + 4.832 = 8.512

B = 8.512 / 2 = 4.256 \mu C

A = 3.68 - B = 3.68 - 4.256 = -0.576 \mu C

6 0
2 years ago
While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6
Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

Let the distance from Boston to Hartford be S.

Person A drives at a constant speed of 55 mph for the entire trip,

Time taken by person A

             t_A=\frac{S}{55}

Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.

Time taken by person B

            t_B=\frac{\frac{S}{2}}{65}+\frac{\frac{S}{2}}{45}=\frac{S}{130}+\frac{S}{90}=\frac{220S}{130\times 90}=\frac{11S}{585}

Ratio of time of arrival of B to A

                      \frac{t_B}{t_A}=\frac{\frac{11S}{585}}{\frac{S}{55}}=\frac{121}{117}=1.034

B will take 1.034 times the time of A from Boston to Hartford.

8 0
2 years ago
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