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natulia [17]
3 years ago
6

A 7.00- kg bowling ball moves at 3.00 m/s. How fast musta

Physics
1 answer:
ratelena [41]3 years ago
6 0

Answer:

Ping Pong ball move at velocity 160.4 m/s

Explanation:

given data

mass m1 = 7 kg

velocity v1 = 3 m/s

mas m2 = 2.45 g = 2.45 × 10^{-3} kg

same kinetic energy

to find out

How fast Ping Pong ball move (v2)

solution

we know same KE

so

\frac{1}{2}* m1* v1^{2} = \frac{1}{2}* m2* v2^{2}   ...........1

so v2 will be

v2 = \sqrt{\frac{m1*v1^2}{m2} }    .............2

put here value in equation 2 we get v2

v2 = \sqrt{\frac{7*3^2}{2.45*10^{-3}}}

v2 = 160.4

so Ping Pong ball move at velocity 160.4 m/s

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They are all coverd in water 24/7 they never clear up
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You multiply force times friction
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A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th
BartSMP [9]

<u>Answer:</u>

The height of ramp = 124.694 m

<u>Explanation:</u>

Using second equation of motion,

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

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

156.3 = 31\times t + 0

t = \frac{156.3}{31 }

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 m/s^2

Substituting in h = ut + \frac{1}{2}gt^2

h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2

h = 124.694 m

So height of ramp = 124.694 m

3 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

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2 years ago
Match the term with each description:
Alex_Xolod [135]

Answer:

An acid is a compound that increases hydrogen ions (H+) when it is dissolved in a solution.

all I got

4 0
3 years ago
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