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2 years ago

Which is at rest relative to a seated passenger in a moving car?

Physics

2 answers:

AnnyKZ [126]2 years ago

5 0

THAT'S EASY BRO! the answer is the Seat!

Alisiya [41]2 years ago

5 0

Answer:

Explanation:

When a person is sitting in a moving car, he is in motion relative to the ground but he is at rest relative to the car.

Everything in the car is at rest relative to the person and the person is also at rest relative to other things in the car.

Thus, everything in the car is at rest relative to the person which is sitting in the car.

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A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts

galina1969 [7]

Answer:

$\mu_s = 0.62$

$\mu_k = 0.415$

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

$x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2$

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

$F = ma\\F = 1.7m$

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

$F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415$

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

$mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62$

5 0

2 years ago

Two balls are at the same height and released at the same time. One ball is dropped and hits the ground 5s later. The 2nd ball i

ASHA 777 [7]

Answer:

The second ball hits at the same time.

Distance travelled is 65 meters

Explanation:

5 0

1 year ago

What is the effect of load resistor on ripple factor

Zolol [24]

ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)

7 0

2 years ago

As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? As a rock sinks deepe

Sidana [21]

Answer:

It remains constant

Explanation:

As we know that buoyant force on an object given as

Fb = ρ Vd g

ρ= Density of fluid

Vd=Volume displace by body

g=10 m/s²

Fb =buoyant force

So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.

So when rock gets deeper and deeper the buoyant force will remain constant.

It remains constant

3 0

2 years ago

A car is traveling with a velocity of 24.4 m/s. It accelerates at a constant rate of 3.2m/s2. If the acceleration lasts for 5.6

larisa86 [58]

Answer: 42.32 ms-1

Explanation:

v = u+ at

v = 24.4 + ( 3.2×5.6)

v = 42.32 ms-1

6 0

2 years ago