A car is traveling at 21m/s. It accelerates at 1.4m/s2 for 11s. How fast is the car moving after the acceleration?

What are the applications of pascal's principle

Murrr4er [49]

Explanation:

The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.
Hydraulic jack- used in the braking system of cars.
use of a straw- to suck fluids, which goes because of air pressure.

<h3>The question simply asks, where pressure can be applied. There are many others, such as lift pump.</h3>

5 0

2 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

3 years ago

4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer

Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

3 0

2 years ago

If the period of the pendulum is tripled, what is the length of the string increased by?

Nitella [24]

The period of the pendulum doesn't determine the length of the string.
It's the other way around.

The period of the pendulum is proportional to the square root of its length.
So if you want to triple the period, you have to make the string nine times
as long as it is now.

7 0

2 years ago

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha

larisa86 [58]

Explanation:

Below is an attachment containing the solution.

7 0

3 years ago