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sergey [27]
3 years ago
8

An unknown material has a mass of 0.447 kg, and its temperature increases by 2.87°C when 943 J of heat are added. What is the sp

ecific heat of the material? (Unit=J(kg*C))
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

735 J/kg/C

Explanation:

Q = mcT

943 = (0.447)( c )(2.87)

1.28289c = 943

c = <u>7</u><u>3</u><u>5</u><u> </u><u>J</u><u>/</u><u>k</u><u>g</u><u>/</u><u>C</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u>

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Answer:The net force on the block is zero.

Explanation:

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Block is being pulled upward along an inclined surface at a constant speed

As speed is constant and moved in a straight line along the plane therefore its velocity is also constant .

and change in velocity is equal to acceleration therefore acceleration is zero here i.e. net force is zero acting on the body.

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Two cylindrical resistors are made of the same material and have the same resistance. The resistors, R1 and R2, have different r
Paladinen [302]

Answer:

Option d is correct.

Explanation:

We know , resistance of a body is directly proportional to its length and inversely proportional to its area.

R=\dfrac{\rho\ L}{A}=\dfrac{\rho\ L}{\pi r^2}      ( Here, \rho is constant dependent on object material )

Writing R_1 \ and\ R_2 also :

R_1=\dfrac{\rho\ L_1}{\pi r_1^2}\ , \  R_2=\dfrac{\rho\ L_2}{\pi r_2^2}      ( since they are of same material therefore, \rho is same.)

Now , if r_2=2r_1 \ and \ 4L_1=L_2.

Then R_1=R_2

Therefore, option d. is correct.

Hence, this is the required solution.

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4 0
3 years ago
Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00m during the first second.
Alexus [3.1K]

Answer:

a) 9.00 m b) 6.00 m/s  c) 12.00 m/s

Explanation:

a) If the acceleration is constant, and we know that the displacement during the first second was 3.00 m, as the boulder (assumed that we can treat it as a point mass) started from rest, we can say the following:

Δx = \frac{1}{2}*a*t^{2} = 3.00 m

As t = 1 s, replacing in the expression above, and solving for a, we have:

a = \frac{2*3.00m}{1s2} = 6.00 m/s²

In order to know how far it travels during the second second, we need to know the value of the speed after the first second, as it is the initial velocity when the second second begins:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

The total displacement, during the second second, will be as follows:

Δx = v₀*t + \frac{1}{2}*a*t^{2} = 6.00m/s*1s +\frac{1}{2}*6.00 m/s2*1s^{2}  = 9.00 m

⇒ Δx = 9.00 m

b) At the end of the first second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

c) At the end of the second second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*2s = 12.00 m/s

3 0
3 years ago
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