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aalyn [17]
1 year ago
10

The cricket ball has a mass of 0.16kg and it hits the bat with a speed of 25 m/s. After being in contact with the bat for 0.0013

s, the ball rebounds with a speed of 22 m/s in the direction exactly opposite to its original direction.
A) state the difference between speed and velocity
B) calculate.
I) the change in velocity of the cricket ball
Ii) the average acceleration of the ball whilst it is in contact with tge bat
Iii) the average force exerted on the ball by the bat​
Physics
1 answer:
zavuch27 [327]1 year ago
7 0

change in velocity is 3m/s,the force exterted by bat on ball is 4.8N and the acceleration is 30m/s².

<h3>Give  some difference between velocity and speed.</h3>

1) Speed is scalar quantity but velocity is vector quantity.

2) Speed is distance followed by an individual with respect to  time and velocity is displacement with respect to time .

1) final velocity- intial velocity

 25-22= 3m/s

2) The force is

F= ma

 . 0.16×30= 4.8N

to learn more about speed and velocity click brainly.com/question/19425181

#SPJ9

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At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar
swat32

Answer:

C=1,25\cdot 10^{5} kJ/^{\circ}C

Explanation:

First of all let's define the specific molar heat capacity.

C = \frac{-Q}{n\cdot \Delta T} (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system  

Now, we can find n with the molar mass (M) the mass of the compound (m).

n=\frac{m}{M}=6.95\cdot 10^{-3} moles      

Using (1) we have:

C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}

C=1,25\cdot 10^{5} kJ/^{\circ}C

I hope it helps!

6 0
3 years ago
A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at
Naddika [18.5K]

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

4 0
3 years ago
) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar
ZanzabumX [31]
Momentum would be the same before and after the collision
 Before the collision:
 Momentum of the single cart: 1 * 0.50 = 0.50
 After the collision
 velocity = 0.25m / s
 1 * 0.25 + 1 * 0.25 =
 0.25 * (1 + 1) =
 0.25 * 2 =
 0.50
 Now new momentum will be 0.5
 answer
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4 0
3 years ago
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A bell rings at a frequency of 75hz on a warm 25 degree evening. calculate the...
allochka39001 [22]

Answer:

Explanation:

We need 2 different equations for this problem: first the velocity of sound equation, then the frequency of the sound equation.

The velocity of sound is found in:

v = 331.5 + .606T

We need to find that first in order to fill it into the frequency equation which is

f=\frac{v}{\lambda} where v is the velocity we will find the part a, f is frequency and lambda is the wavelength. Starting with the velocity of the sound:

v = 331.5 + .606(25) and

v = 331.5 + 15 and rounding correctly using the rules for sig fig when adding:

v = 347 m/s

Filling that into the frequency equation:

75=\frac{347}{\lambda} and

\lambda=\frac{347}{75} so

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7 0
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Equilibrium can occur only if
sergeinik [125]

Answer:

Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. All reactant and product concentrations are constant at equilibrium.

Explanation:

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