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Natalija [7]
3 years ago
8

A car with a mass of 1500 kg is pulled by a rope that is horizontal to the ground. The tension in the rope is 2000 N and a frict

ion force of 350 N opposes the car's motion. What is the magnitude of the car's acceleration?
Physics
1 answer:
Tanzania [10]3 years ago
7 0

Answer:

Explanation:

Assuming the ground is level as well.

F = ma

a = F/m

a = (2000 - 350) / 1500

a = 1.1 m/s²

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A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal
Elina [12.6K]

Complete question:

A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.

Answer:

The mass of the air is 5.32 kg

The specific volume is 5.52 x 10⁻³ m³/kg

Explanation:

Given;

total volume of the container, V_t = 5 m³

mass of granite, m_g = 900 kg

density of granite, \rho _g = 2,400 kg/m³

density of air, \rho_a = 1.15 kg/m³

The volume of the granite is calculated as;

V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3

The volume of air is calculated as;

V_a = V_t - V_g\\\\V_a = 5 \ m^3  \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3

The mass of the air is calculated as;

m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg

The specific volume is calculated as;

V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg

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Radiation can carry thermal energy.
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The fact that electric charges return to the source of the current is an example of:
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The law of conservation of charge.

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A car traveling at 15 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration ?
Nataliya [291]

Answer:

-3m {s}^{-2}

Explanation:

The initial velocity, u, of the car=15m/s

The final velocity, v, of the car =0m/s

Time, t, taken for the car to come to a stop=5s

Acceleration is calculated by,

a= \frac{v - u}{t}

By substitution,

a= \frac{0- 15}{5}

a= \frac{- 15}{5}

a = -3 {ms}^{ - 2}

The negative sign implies that the car has decelerated.

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