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N76 [4]
3 years ago
15

A 1200-kg car pushes a 2100-kg truck that has a dead battery to the right. When the driver steps on the accelerator, the drive w

heels of the car push against the ground with a force of 4500 N. Which, if any, of the following gives a correct acceleration constraint for this problem? The variable ac is the acceleration of the car, and the variable t is the acceleration of the truck.
a. ac=at
b. ac=−at
c. ac=21001200at
d. ac=12002100at
e. There are no acceleration constraints for this problem.
Physics
1 answer:
worty [1.4K]3 years ago
8 0

Answer:

Option A: a_c = a_t

Explanation:

Let's denote the force acting on the car due to the truck as F_ct

Similarly, the force acting on the truck due to the car is denoted as F_tc.

Now, F_ct = m_c × a_c

Where;

m_c is mass of car

a_c is acceleration of car

Also, F_tc = m_t × a_t

Where;

m_c is mass of truck

a_c is acceleration of truck

Now, since the car is the one pushing the truck, from Newton's laws, we can say that they stick together and move in the same direction at the same acceleration.

Thus, the truck will move with an acceleration equal to that of the car pushing it.

Thus; a_c = a_t

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8 0
3 years ago
An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from
uranmaximum [27]

Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

7 0
3 years ago
A 10 kg package is delivered to your house.
kolbaska11 [484]
In ur explanation make sure to use the terms
7 0
3 years ago
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s
AlexFokin [52]

Answer:

K=0.023J

Explanation:

From the question we are told that:

Mass m=0.15

Velocity v=0.5m/s

Angular Velocity \omega=8.4rad/s

Generally the equation for Kinetic Energy is mathematically given by

K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)

K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))

K=0.023J

8 0
3 years ago
A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu
kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

\omega f = \omega o-\alpha*t

Solving for α:

\alpha=\frac{\omega o-\omega f}{t}

where ωo = 600*2*π/60;   ωf = 0;    t=10s

\alpha=6.283rad/s^2

The sum of torque is:

\tau=I*\alpha

\tau=M*R^2/2*\alpha

\tau=0.060 N.m

8 0
3 years ago
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