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Rzqust [24]
3 years ago
15

A ball is thrown with 50J of kinetic energy, it hits a target which moves with 30J of kinetic energy, how much energy goes to th

e thermal store of the surroundings?
Physics
1 answer:
djverab [1.8K]3 years ago
3 0

Answer:

The energy that will go will for thermal store of the surroundings is 20 J.

Explanation:

Given;

kinetic of the thrown ball, K.E₁ = 50 J

kinetic energy used to move the target, K.E₂ = 30 J

The excess energy that will go will for thermal store of the surroundings;

ΔK.E = K.E₁ - K.E₂

ΔK.E = 50J - 30J

ΔK.E = 20 J

Therefore, the energy that will go will for thermal store of the surroundings is 20 J.

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What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters
Volgvan
A_central = v^2/r = (19)^2/10 = 36.1 m/s^2
7 0
3 years ago
Read 2 more answers
A 55.4 g sample of water at 99.61 °C is placed in a constant pressure calorimeter. Then, 23.4 g of zinc metal at 21.6 °C is adde
Zolol [24]

Answer:

The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C

Explanation:

From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture

Heat loss by water = Heat gain by zinc metal

m1C1(T1 - T3) = m2C2(T3 - T2)

m1 is mass of water = 55.4 g

C1 is specific heat capacity of water = 4.2 J/g.°C

m2 is mass of zinc metal = 23.4 g

C2 is specific heat capacity of zinc metal

T1 is the initial temperature of water = 99.61 °C

T2 is the initial temperature of zinc metal = 21.6 °C

T3 is the final temperature of the mixture = 96.4 °C

55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)

746.9028 = 1750.32C2

C2 = 746.9028/1750.32 = 0.427 J/g.°C

3 0
3 years ago
Why is electrostatic force able to act at a distance ​
mote1985 [20]
In Electrostatics the electrical force between Two charged objects is inversely Related to the distance of separation between the two objects .
7 0
3 years ago
A body travels 30m in 5s, 45m in 7s and then 65m in the last 5s. Find the average speed of the body
JulijaS [17]
We can find the average speed of the body by finding the total distance covered, and then dividing it by the total time of the motion.

The total distance covered is:
S=30 m + 45m+65m=140 m

while the total time of the motion is
t=5 s+7s+5 s=17 s

So, the average speed of the body is:
v= \frac{S}{t}= \frac{140 m}{17 s}=8.24 m/s
8 0
3 years ago
What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28m/s
nika2105 [10]

Answer:

<em>An average net force of 5200 N is needed to stop the car</em>

Explanation:

<u>Cinematics and Dynamics</u>

Cinematics describes the variables involved in the movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time, and displacement (or the scalar equivalent, distance).

The acceleration can be calculated by:

\displaystyle a=\frac{v_f-v_o}{t}

The initial speed is vo=28 m/s, it stops (vf=0) in t=10.5 seconds, thus the acceleration is:

\displaystyle a=\frac{0-28}{10.5}

a = -2.67~m/s^2

The acceleration is negative because the car loses speed.

Knowing the mass of the car m=1950 Kg, we can calculate the net force required to stop the car by using the formula:

F = m.a =1950*2.67

F = 5200 N

An average net force of 5200 N is needed to stop the car

7 0
3 years ago
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