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mamaluj [8]
3 years ago
13

A 2.0 Kg box is being pulled up a ramp with a force of 25.0 Newtons. The ramp has an incline of 30 degrees. If the coefficient o

f friction is 0.752, find the acceleration of the block up the ramp. You may round your answer to the nearest tenth where necessary!!!
Physics
1 answer:
nydimaria [60]3 years ago
5 0

Answer:

The acceleration of the block is 1.21 m/s².

Explanation:

Given that,

Mass = 2.0 kg

Force = 25.0 n

Angle = 30°

Coefficient of friction = 0.752

We need to calculate the acceleration of the block

Using balance equation

F-(mg\sin\theta+\mu mg\cos\theta)=ma

Put the value into the formula

25-(2.0\times9.8\sin30+0.752\times2.0\times9.8\cos30)=2.0a

a=\dfrac{25-(2.0\times9.8\sin30+0.752\times2.0\times9.8\cos30)}{2.0}

a=1.21\ m/s^{2}

Hence, The acceleration of the block is 1.21 m/s².

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4 0
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A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
3 years ago
A block of ice (m = 9 kg) at a temperature of T1 = 0 degrees C is placed out in the sun until it melts, and the temperature of t
jonny [76]

Answer:

a) An expression for the amount of energy, E_m, needed to melt the ice into water.

(E_m) = (m × Lf)

b) An expression for the total amount of energy, E_tot, to melt the ice and then bring the water to T2

(Total heat) = (m × Lf) + mc (T2 - T1)

c) 3,646,458 J = 3646.46 kJ

Explanation:

a) When a pure body changes its phase at meltimgbor boiling point, it does so at a constant temperature. When a pure body melts, the amount of heat responsible for this change is just given by a product od the mass of the body and the body's heat of fusion.

(E_m) = (m × Lf)

b) The Heat required to raise the temperature of a body from one temperature to another is given by the product of the mass of the body, its specific heat capacity and the temperature difference between the final point and the starting point.

(E_2) = mcΔT = mc (T2 - T1)

Total heat required to melt the ice at T1 = 0 and raise the temperature of the resulting water to T2 is then a sum of (E_m) + (E_2)

(Total heat) = (m × Lf) + mc (T2 - T1)

c) What is the energy in Joules?

(Total heat) = (m × Lf) + mc (T2 - T1)

m = mass of ice = resulting mass of water = 9 kg

Lf = latent heat of fusion = 334000 J/kg

c = Specific heat capacity of water = 4186 J/kg.K

T2 = final temperature of the water = 17°C

T1 = Initial temperature of the water = 0°C

Note that the units of temperature difference is the same for K and °C

(Total heat) = (m × Lf) + mc (T2 - T1)

Q = (9 × 334000) + [9 × 4186 × (17 - 0)]

Q = 3,006,000 + 640,458 = 3,646,458 J = 3646.46 kJ

Hope this Helps!!!

7 0
4 years ago
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ValentinkaMS [17]

Answer:

45 s .

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displacement s = ?

acceleration a = 1 m /s²

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initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

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