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exis [7]
3 years ago
10

a 100gm copper block is heated in boiling water for 10min and then it is dropped into 150gm of water at 30 C in a 200gm calorime

ter.if the temperature of water is raised to 33.6 C . determine the specific heat of material of calorimeter?
Physics
1 answer:
Gekata [30.6K]3 years ago
8 0
Don't ask me son i don't know
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A car starts from rest and accelerates uniformly over a time of 7.25 seconds for a distance of 210 m. Determine the acceleration
lara31 [8.8K]

Answer:

Acceleration is 7.990487515m/s²

Initial velocity is 0m.s

Explanation:

s=ut+(1/2)at²

210=0(7.25)+(1/2)a(7.25²)

210=26.28125a

∴a=7.990487515m/s²

'Vi' or 'u' is the inital speed. Since it starts from rest, this equals 0.

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Is diet Pepsi heterogeneous or is it homogeneous
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Heterogeneous because it has less amount of sugar than the regular Pepsi has
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3 years ago
A length change 0.08 m will occur for an object that is L= 56 m long. If the coefficient of thermal expansion is5.3 x 10 /C and
ArbitrLikvidat [17]

Answer:

Increase in temperature =  269.54 °C

Explanation:

We have equation for thermal expansion

          ΔL = LαΔT

Change in length, ΔL = 0.08 m

Length, L = 56 m

Coefficient of thermal expansion, α = 5.3 x 10⁻⁶ °C⁻1

Change in temperature, ΔT = T - 253

Substituting

          0.08 = 56 x 5.3 x 10⁻⁶ x (T - 253)

         (T - 253) = 269.54

           T = 522.54 °C

Increase in temperature =  269.54 °C      

4 0
3 years ago
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0
3 years ago
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