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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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3 ways pollution gets into neighborhoods
SashulF [63]

Answer:

1. Leaving trash on the ground and not picking it up.

2. Polluting the lake, if your neighborhood has one.

3. Outside people come and dump their waste to your neighborhood.

Hope this helps!

7 0
3 years ago
Which best describes friction?
tangare [24]

Answer:

a constabt force that acts on object that rub together

Explanation:

It is because Friction is the rubbing of one body against another body

5 0
2 years ago
These waves are most harmful for living things.
scZoUnD [109]

Explanation:

G. gamma rays because they are produced by the hottest and most energetic objects in the universe, such as neutron stars and pulsars, supernova explosions, and regions around black holes.

hope this helps you

have a nice day:)

5 0
3 years ago
A human subject with mass 96 kg, body specific heat 3500 Jkg^-1K^-1 skin temperature 34.5 degrees Celsius, surface area 1.5m^2 a
AlekseyPX

rate of heat radiation by the body is given by

\frac{dQ}{dt} = \sigma e A(T^4 - T_s^4)

here we know that

\sigma = 5.67 \times 10^{-8}

e = 0.7

A = 1.5 m^2

T = 34.5 + 273 = 307.5 k

T_s = 21 + 273 = 294 k

now from above formula rate of heat dissipation

\frac{dQ}{dt} = (5.67 \times 10^{-8}}(0.7)(1.5)(307.5^4 - 294^4)

\frac{dQ}{dt} = 87.5

now we know that

Q = ms \Delta T

from above equation

\frac{dQ}{dt} = ms\frac{dT}{dt}

now we have

m s \frac{dT}{dt} = 87.5

here we have

m = 96 kg

s = 3500

96 \times 3500 \times \frac{dT}{dt} = 87.5

\frac{dT}{dt} = (2.6 \times 10^{-4}) \: ^0C/s

\frac{dT}{dt} = 0.94 ^0 C/hour

7 0
3 years ago
a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of
NeTakaya

Answer:

The acceleration of the object is 5.57\ m/s^2.

Explanation:

Given that,

Mass of the block, m = 2.6 kg

Spring constant of the spring, k = 126 N/m

At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.

When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :

kx=ma

a is acceleration of the object

a=\dfrac{kx}{m}\\\\a=\dfrac{126\times 0.115}{2.6}\\\\a=5.57\ m/s^2

So, the acceleration of the object is 5.57\ m/s^2.

8 0
3 years ago
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