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Bad White [126]
3 years ago
5

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

your feet and the ice. A friend throws you a ball of mass 0.400 kg that is traveling horizontally at 11.5 m/s. Your mass is 75.0 kg.
(A) If you catch the ball, with what speed do you and the ball move afterwards?
(B) If the ball hits you and bounces off your chest, so that afterwards it is moving horizontally at 7.80m/s in the opposite direction, what is your speed after the collision?
Physics
1 answer:
9966 [12]3 years ago
3 0

Answer:

(a) 0.061 m/s

(b) 0.103 m/s

Explanation:

From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum

Momentum, p=mv where m is the mass and v is the velocity

m_1v_1=(m_1+m_2)v_c where v_c is the common velocity, v_1 is the velocity of the ball m_1 and m_2 are masses of the ball and person respectively

Substituting the given values then

0.400\times 11.5 = (0.400+75)v_c\\v_c=0.061007958\approx 0.061 m/s

(b)

Momentum, p=mv where m is the mass and v is the velocity

m_1v_1=m_1v_2+m_2v_3

v_1 is the velocity of the ball , v_2 is the velocity of ball afterwards and v_3 is your speed, m_1 and m_2 are masses of the ball and person respectively. Since it bounces back, we give it a negative value hence

0.400\times 11.5= 0.4\times -7.8+75v_3\\v_3=0.102933333\approx 0.103 m/s

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In addition to average weather conditions, climatological data also describes annual variations and fluctuations of temperature,
drek231 [11]

<u>Answer:</u>

In addition to average weather conditions, climatological data also describes annual variations and fluctuations of temperature, precipitation, wind speed and other variables.

<u>Explanation</u>:

A lot many observations are made all around the world regarding the weather each day. These observations and analysis are done by humans as well as automated instruments. The weather data is collected each day all year and any inaccuracies and discrepancies are checked and rectified. The results are later then presented as the climate data. There are various factors that are taken into consideration while determining the climate of a region. Apart from the factors that are already mentioned, wind speed is also one of the other variables.

8 0
3 years ago
You drop a batitin a stationary elevator and the ball hits the floor in o 50 s. How long does it take for the ball to hit the fl
solmaris [256]

Answer:

option (a) 0.61 s

Explanation:

Given;

Time taken by the ball to reach the ground  = 0.50 s

Let us first calculate the distance through which the ball falls on the ground

from the Newton's equation of motion, we have

s=ut+\frac{1}{2}at^2

where,  

s is the distance

a is the acceleration

t is the time

here it is the case of free fall

thus, a = g = acceleration due to gravity

u =  initial speed of the ball = 0

on substituting the values, we get

s=0\times 0.50+\frac{1}{2}\times9.8\times0.50^2

or

s = 1.225 m

Now,

when the elevator is moving up with speed of 1.0 m/s

the initial speed of the ball = -1.0 m/s   (as the elevator is moving in upward direction)

thus , we have

s=ut+\frac{1}{2}at^2

or

1.225=-1.0\times\ t+\frac{1}{2}\times9.8t^2

or

4.9t^2 - t  - 1.225 = 0

or

t = 0.612 s

hence, the correct answer is option (a) 0.61 s

4 0
4 years ago
an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying a
mart [117]

Answer:

331.7m/s

Explanation:

Given parameters:

Initial velocity  = 100m/s

Acceleration  = 50m/s²

Distance  = 1km   = 1000m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we have to apply the right motion equation shown below;

     v²  = u²  + 2aS

 v is the final velocity

 u is the initial velocity

 a is the acceleration

 S is the distance

 Now insert the parameters and solve;

     v² = 100² + (2 x 50 x 1000)

     v² = 110000

     v = √110000  = 331.7m/s

4 0
3 years ago
I've got an energy and work problem. The premise of the problem is:
Alenkasestr [34]
Refer to the diagram shown below.

μ =  the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
    The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
    mg sin(30) = 0.5mg. 
    The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
     This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
    The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
     F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ) 

6. The work done on the crate by the total force is
    d*(F - 0.5mg - 0.866μmg)

7 0
4 years ago
A 1.97-pF capacitor is connected to a 9.0-V battery and fully charged. How many electrons did the battery transfer from one capa
Dovator [93]

Answer:

1.11×10⁸  Electrons.

Explanation:

Applying,

Q = CV..................... Equation 1

Where Q = Total charge on the capacitor, C = Capacitance of the capacitor, V = Voltage of the battery.

Given: C = 1.97 pF = 1.97×10⁻¹² F, V = 9.0 V

Substitute this value into equation 1

Q = 1.97×9×10⁻¹²

Q = 1.773×10⁻¹¹ C.

If the charge on one electron = 1.602×10⁻¹⁹ C.

Therefore number of electron = 1.773×10⁻¹¹ /1.602×10⁻¹⁹

Number of electron = 1.11×10⁸

8 0
3 years ago
Read 2 more answers
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