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ss7ja [257]
3 years ago
11

What is the acceleration of a 45 kg mass pushed with 15 N of force? What is the acceleration if the force changes to 25 N? What

if the mass changes to 70 kg?
Physics
2 answers:
Rama09 [41]3 years ago
8 0
F = ma
Rearrange this so acceleration is the subject:
A = f/m

Now input your values into this equation for the first question :
A = f/m
A = 15/45
A = 0.33333333 m/s

Then using this same equation change the force of 15 to 25 :
A = f/m
A = 25/45
A = 0.55555556 m/s

For the last question keep the force of 15N but change the mass to 70kg into the same equation :
A = f/m
A = 15/70
A = 0.21 m/s (rounded)
Contact [7]3 years ago
6 0

Answer:

Part a)

a = \frac{1}{3} m/s^2

Part b)

a = \frac{5}{9} m/s^2

Part c)

a = \frac{3}{14} m/s^2

Explanation:

As per Newton's II law we know that

F = ma

Now we know that

m = 45 kg

F = 15 N

so we will have

15 = 45 a

a = \frac{15}{45} = \frac{1}{3} m/s^2

Now we have

m = 45 kg

F = 25 N

now again from Newton's Law

25 = 45 a

a = \frac{25}{45} m/s^2

a = \frac{5}{9} m/s^2

Now if mass is changed to 70 kg

again we have

15 = 70 a

a = \frac{15}{70}

a = \frac{3}{14} m/s^2

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4 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
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<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

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\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

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m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

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What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

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