Could you give me 4 examples of acceleration??

1 answer:

5 0

Certainly.

1). A driver presses the gas and the car speeds up.

2). A driver presses the brake and the car slows down.

3). A driver turns the steering wheel slightly. The car's speed doesn't change
as it goes around the curve.

4). A stone is tossed into the air. Gravity accelerates it from the moment it
leaves the thrower's hand until it hits the ground.

5). A new TV satellite is successfully launched and placed in a perfect circular
orbit around the Earth. The satellite's motion is accelerated forever, 24/7/365 .

"Acceleration" means any change in the speed or direction of motion. Any
object that's not moving at constant speed in a straight line is accelerated.

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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

8 0

3 years ago

What type of noise is most difficult to remove from a digital signal? why?

Lilit [14]

Interference if the hardest to remove due to the superposition of the frequencies of the radio waves

8 0

3 years ago

A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force

melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

$\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}$

in which $\small \Delta (mV)$ , $\small \Delta t$ represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by $\small F_1$ , the following manipulation confirms the answer to this question.

$\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}$