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3 years ago

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What is the final temperature of a block if it’s thermal energy increases by 2,130 that has a mass of 50kg, a specific heat of 7

10J/kgC and is at a temperature of 20 degrees Celsius.

1 answer:

BartSMP [9]3 years ago

4 0

Answer:

I dont know

Explanation:

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20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?

Tresset [83]

Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance
so Work done = 0.085 x 20g x 28
= 466.956 J

7 0

3 years ago

What is the relationship between force in velocity selector in a bain bridge.

kifflom [539]

Answer:

In a velocity selector, there are two forces namely;

» Electric field Intensity

» Magnetic field density

<u>Relationship</u><u>:</u>

$E _{e}= B$

E is the electric field intensity

B is the magnetic flux density

4 0

3 years ago

A worker pushes horizontally on a large crate with a force of 200 N, and the crate is moved 3.5 m. How much work was done in Jou

sattari [20]

Work = (force) x (distance) =

(200 N) x (3.5 m) = <em>700 joules</em>

6 0

3 years ago

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

$a = 11.8 m/s^2$

8 0

3 years ago

A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the we

Alex_Xolod [135]

Answer:

w = 2w₀ the angular velocity of man doubles

Explanation:

In this exercise, releasing the weights reduces the moment of inertia

I= I₀ / 2

Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved

L₀ = L

I₀ w₀ = I w

I₀ w₀ = I₀ / 2 w

w = 2w₀

therefore the angular velocity of man doubles

7 0

3 years ago