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vlada-n [284]
3 years ago
7

B

Physics
1 answer:
ololo11 [35]3 years ago
3 0

Answer:

Thermal capacity of the object = 720 j/°C

Explanation:

Given:

Specific heat capacity of metal = 360J/(kg°C)

Mass of given object = 2 kg

Find:

Thermal capacity of the object

Computation:

Thermal capacity = Specific heat x Mass

Thermal capacity of the object = Specific heat capacity of metal x Mass of given object

Thermal capacity of the object = 360 x 2

Thermal capacity of the object = 720 j/°C

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On the surface of Earth, the force of gravity acting on one kilogram is: *
Murrr4er [49]

Your answer would be:

Option (B) : 2.2 Pounds

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4 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
What role does actin and myosin play in the process of muscle contraction?
Margarita [4]

Answer:

Muscle contraction thus results from an interaction between the actin and myosin filaments that generates their movement relative to one another. The molecular basis for this interaction is the binding of myosin to actin filaments, allowing myosin to function as a motor that drives filament sliding.

4 0
2 years ago
Select the correct answer.
VladimirAG [237]

Answer:

It's effective temperature.

Explanation:

8 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
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