All categories

2 years ago

A 1,000 kg car is moving at 20 m/s, and slams into a building before coming to a halt.

Physics

1 answer:

ale4655 [162]2 years ago

7 0

Hi there!

We can calculate linear momentum using the following:
$\large\boxed{p = m v}$

p = linear momentum (kgm/s)
m = mass (kg)
v = velocity (m/s)

Calculate:

$p = 1000 * 20 = 20000 kg\frac{m}{s}$

Now, we can relate force, time, and momentum with the following:
$\large\boxed{I = Ft}\\\\$

I = Impulse (kgm/s)
F = Force (N)
t = time (s)

Rearrange to solve for force:
$F = \frac{I}{t}$

The impulse is equal to the change in momentum. Since the car comes to a halt, all of its momentum is lost, so:

$I = \Delta p = p_f - p_i = 0 - 20000 = -20000$

Solve:
$F = \frac{-20000}{0.5} = \boxed{-40000N}$

**Negative force since the positive direction is towards the wall, and the negative direction is away from the wall.

You might be interested in

Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m

marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0

4 years ago

12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car c

Triss [41]

Answer:

22.5 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Time (t) = 1.5 s

Final velocity (v) = 0 m/s

Distance (s) =?

The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:

s = (u + v)t/2

s = (30 + 0)1.5 / 2

s = (30 × 1.5) / 2

s = 45 / 2

s = 22.5 m

Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.

4 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago

In a tug of war between Mrs. Brenneman and Mr. Schroedl seems at a standstill. Then Mrs. Brenneman tugs hard giving a force of 4

sergiy2304 [10]

He feels a 10 N to the left force moves. Yes ,he moves.

4 0

3 years ago

13.

lisov135 [29]

The net torque on the seesaw is 294 Nm.

<h3>What is torque?</h3>

Torque is the force that tends to rotate the body to which it was applied.

To calculate the net torque, we use the formula below

Formula:

τ = mgd-m'gd'........... Equation 1

Where:

τ = Net torque

m = Jenny's mass
m' = Tom's mass
d = Jenny's distance from the pivot
d' = Tom's distance from the pivot
g = acceleration due to gravity

From the question,

Given:

m = 40 kg
m' = 30 kg
d = 1.5 m
d' = 1 m
g = 9.8 m/s²

Substitute these values into equation 1

τ = (40×1.5×9.8)-(30×1×9.8)
τ = 588-294
τ = 294 Nm

Hence, The net torque on the seesaw is 294 Nm.

Learn more about torque here: brainly.com/question/14839816

5 0

3 years ago