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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

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Larry Finkelstein, Norman Fischer, and Cassius Schwartz have been overlooked, in my opinion.
8 0
2 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
2 years ago
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Fris
Usimov [2.4K]

Answer:

Torque τ =w ×0 = 0

Explanation:

We know that the torque is given by the product of the force and perpendicular distance between the force and the axis.

Here the gravity force act at the center and the rotational axis is also passing through the center.

Therefore the perpendicular distance between the force and the rotational axis would be zero.

Hence the torque will be

Torque = Force × Perpendicular distance

Torque = mg×0 = 0

Therefore the torque would be zero.

7 0
2 years ago
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0
2 years ago
A wagon full of manure accidentally rolls down a driveway for 5.0m while a person pushes against the wagon with a force of 420 N
Cerrena [4.2K]

Answer:

2100 J

Explanation:

Parameters given:

Force acting on the object, F = 420 N

Distance moved by object, d = 5m

The change in kinetic energy of an object is equal to the work done by a force acting on the object:

W = F * d

∆KE = F * d

∆KE = 420 * 5

∆KE = 2100 J

8 0
2 years ago
Read 2 more answers
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