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bogdanovich [222]
3 years ago
10

A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What is the velocity of the bowling ball?

Physics
1 answer:
KIM [24]3 years ago
7 0

Answer:

6.19m/s

Explanation:

Given parameters:

Mass of bowling ball  = 35.2kg

Momentum  = 218kgm/s

Unknown:

Velocity of the bowling ball  = ?

Solution:

Momentum is the quantity of motion a body possess;

    Momentum  = mass x velocity

   218  = 35.2 x velocity

     Velocity  = 6.19m/s

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John wants a new video tape and a new shirt, but he only has $12.00. He buys the video tape and gets $0.52 change. What was his
mestny [16]

Answer:

You have the answer in your comments. I will be copying it so your question doesn't get deleted.

The answers is $0.58

$11.48

the video tape

the new shirt

7 0
2 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.
Lelechka [254]

Answer:

B. 161.5 J

Explanation:

n = Number of moles = 0.37

\Delta T = Rise in the temperature of the oxygen gas = 15 K

Q = heat added in order to raise the temperature

c_{p} = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J

6 0
3 years ago
When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the
o-na [289]
The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.

(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m

Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.

(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ

θ=37.69 degrees

Hopes this helps!

P.S. I hope this is right. If not sorry in advance.
3 0
3 years ago
The change in momentum experienced by a object is equivalent to the...
Papessa [141]

Answer: C (impulse acting on the object)

The momentum is defined as it is the impulse acting on the force . Change in momentum is known as Impulse. Impulse is used to increase or decrease the momentum of object.                                    

From Newtons II law

              F = m. a

                = m. v/t               <em>since a = rate of change of velocity.</em>

<em>              </em>F . t = m . v

F . t is known as impulse momentum

8 0
3 years ago
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