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Amiraneli [1.4K]
3 years ago
8

Two cars collide in a parking lot. Before the collision, their total momentum was 1800 kg m/s to the east. What is their total m

omentum after the collision.
Physics
1 answer:
Pepsi [2]3 years ago
3 0

During collision there is no force on two cars together

So here we can say

F = \frac{\Delta P}{\Delta t}

Since force on the system is zero

0 = \frac{\Delta P}{\Delta t}

\Delta P= 0

P_i = P_f

So after collision momentum will remain the same i.e. 1800 kg m/s

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Hi there!

We can use Newton's Second Law:

\Sigma F = ma

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are given the mass and acceleration, so:

∑F = 20 · 2 = <u>40 N</u>

4 0
2 years ago
Accurate forecasts require careful study of the location, size, movement, and characteristics of large _____.
tankabanditka [31]

Answer: B air masses

3 0
3 years ago
011 10.0 points
Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s

and the time taken to cover the horizontal distance d is

t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

y=u_y t - \frac{1}{2}gt^2

where

u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m

The height of the crossbar is h = 3.05 m, so the ball passes

h' = 5.52- 3.05 = 2.47 m

above the crossbar.

8 0
3 years ago
What is the number of waves that pass a fixed ping in a given amount of time?
Viktor [21]
The number of waves that pass a fixed point in a given amount of time is wave frequency. Wave frequency can be measured by counting the number of crests (high points) of waves that pass the fixed point in 1 second or some other time period. The higher the number is, the greater the frequency of the waves. :)
7 0
2 years ago
A 5.5kg radio is pushed across the table. If the acceleration is 5m/s to the right, find the net force exerted on the radio.
Sholpan [36]

Answer:

Net force exerted on the radio is 27.5 Newton.

Given:

Mass = 5.5 kg

Acceleration = 5 \frac{m}{s^{2} }

To find:

Force exerted on the radio = ?

Formula used:

F = ma

Where F = net force

m = mass

a = acceleration

Solution:

According to Newton's second law of motion,

F = ma

Where F = net force

m = mass

a = acceleration

F = 5.5 × 5

F = 27.5 Newton

Hence, Net force exerted on the radio is 27.5 Newton.


4 0
3 years ago
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