PLEASE HELP!!!

Jack (mass 52.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. he collides with jill (mass 49.0 k

dybincka [34]

We must write down laws of conservation of momenta and energy.
For the law of conservation of momenta will we will use two axes. One will be x-axis that will correspond to the east, and the other one will be y-axis corresponding to the north. Jack will be marked as 1 and Jill will be marked as 2.
Law of conservation of energy:
$\frac{m_1v_1^2}{2}=\frac{m_1v'_1^2}{2}+ \frac{m_2v_2^2}{2}\\
m_2v_2^2=m_1v_1^2-m_1v_1'^2\\
v_2=\sqrt{\frac{m_1v_1^2-m_1v_1'^2}{m_2}$
This will give us Jill's velocity after the colision.
$v_2=6.43\frac{m}{s}$
Law of conservation of momenta:
$x: m_1v_1=m_1v_1'cos(34)+m_2v_2cos(\theta)\\
y: 0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\$
We will use the second equation to get the angle at which the Jill is traveling:
$0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\
m_2v_2sin(\theta)=m_1v_1'sin(34)\\
sin(\theta)=\frac{m_1v_1'sin(34)}{m_2v_2}\\
\theta=sin^{-1}(\frac{m_1v_1'sin(34)}{m_2v_2})$
When we plug all the number we get:
$\theta=27.45^\circ$
Please note that this is the angle below the x-axis.

6 0

3 years ago

A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the

viva [34]

Answer:

d. 6.0 m

Explanation:

Given;

initial velocity of the car, u = 7.0 m/s

distance traveled by the car, d = 1.5 m

Assuming the car to be decelerating at a constant rate when the brakes were applied;

v² = u² + 2(-a)s

v² = u² - 2as

where;

v is the final velocity of the car when it stops

0 = u² - 2as

2as = u²

a = u² / 2s

a = (7)² / (2 x 1.5)

a = 16.333 m/s

When the velocity is 14 m/s

v² = u² - 2as

0 = u² - 2as

2as = u²

s = u² / 2a

s = (14)² / (2 x 16.333)

s = 6.0 m

Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.

The correct option is d

4 0

3 years ago

What is the velocity of a car that travelled a total of 75 km north in 1.5 hrs?

AURORKA [14]

V=d*t
Since d=75km=75000m
t=1.5h=3600*1.5= 5400s
=>V=75000*5400
=405000000m/s=4.05*10^8m/s

3 0

4 years ago

Why is the power of concave lense negative

Travka [436]

Answer:

because the focal length is negative.

Explanation:

Power of a lens (P) is the reciprocal of its focal length (f).

4 0

3 years ago

a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10

TEA [102]

Answer:

Explanation:

Here's the info we have:

initial velocity is 20 m/s;

final velocity is our unknown;

displacement is -10.2 m; and

acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation

v² = v₀² + 2aΔx and filling in accordingly to solve for v:

$v=\sqrt{(20)^2+2(-9.8)(-10.2)}$ Rounding to the correct number of sig fig's to simplify:

$v=\sqrt{400+2.0*10^2}$ to get

v = $\sqrt{600}=20\frac{m}{s}$ If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.

So either 20 m/s or 24 m/s

7 0

3 years ago