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MrMuchimi
3 years ago
6

A 0.05 kg tennis ball moves at a speed of 10 m/s and is struck by a racket causing it to rebound in the

Physics
2 answers:
Ratling [72]3 years ago
5 0
∆=!−! = 0.05 10−−16 =.
= ∆ = ∆ = . −
Lena [83]3 years ago
5 0
F=ma
F=m[(v-u)/t]
Ft=m[v-u]
Ft=mv-mu
Ft=(0.05x10)-(0.05x16)
Ft= -0.3 Ns
Impulse= -0.3 Ns.
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An artillery shell has a mass of 55 kg. The shell is fired from the muzzle of the gun
Evgen [1.6K]

The average force on the shell while it is in the gun barrel is 10,869,833.33N.

<h3>HOW TO CALCULATE FORCE?</h3>

Force applied on an object can be calculated by multiplying the mass of the object by its acceleration.

However, the acceleration needs to be calculated by using the following equation:

v² - u² = 2as

Where;

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770² - 0² = 2 (a)(1.5)

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2 years ago
Which statement is not true about light? 1 point It is a particle and a wave It has mechanical and transverse waves It has mecha
belka [17]

Answer:

Option C

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3 years ago
A car travels 40 miles in 30 minutes.
lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 m/s^2

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles =  40 x 1.60934

so 40 miles  =  64.3738 Km

similarly converting 30 minutes into hours

1 minute = \frac{1}{60}hours

30 minute = \frac{30}{60}hours

30 minute = \frac{1}{2}hours

Now

Average velocity = \frac{Speed}{time}

Substituting Values,

Average velocity = \frac{64.3738}{\frac[1}{2}}

Average velocity = 64.3738 \times 2

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting  velocity to m/s

128.74  \times \frac{5}{18}

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy  K= \frac{1}{2}mv^2

Substituting the values,

K= \frac{1}{2}1500(35.75)^2

K= \frac{1}{2}1500(1278.06)

K= \frac{1500 \times (1278.06)}{2}

K= \frac{1917093.75}{2}

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

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

Substituting the known values,

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

s= (37.75)(15)+\frac{1}{2}a(15)^2

s=566.25+\frac{1}{2}a(225)

s=566.25+\frac{(225a)}{2}-------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

a = \frac{v - u}{t}

Substituting the values

a = \frac{0 - 35.75}{15}

a = \frac{- 35.75}{15}

a= - 2.38 m/s^2----------------------------------------(4)

Now substituting 4 in 3 we get

s=566.25+\frac{(225( - 2.38)}{2}

s=566.25+\frac{-535.5}{2}

s=536.25-267.75

s=268.8m--------------------------------------------------------------(5)

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