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Levart [38]
3 years ago
13

A man is running on the straight road with the uniform velocity of 3 metre per second calculate the acceleration for produced hi

m​
Physics
1 answer:
sp2606 [1]3 years ago
6 0

Theoritically

the body moving with uniform velocity has acceleration zero.

Mathmatically,

u=3m/s

v=3m/s (since body is moving with uniform velocity)

a= v-u/t

3-3/t

0/t

0m/s.s

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Calculate the G.P.E. in joules of a 675-newton climber at the top of a 3,050-meter mountain in Colorado.
almond37 [142]

Answer:2673

Explanation:

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3 years ago
Amy swims 500 m (0.5 Km) in 6 minutes (0.1 hour). What was her speed in Kilometers per hour? A. 8.3 Km/hr B. 5 Km/hr C. 50 Km/hr
kondor19780726 [428]
Speed = Distance ÷ Time so divide .5 km by .1h. .5 km÷.1h=5 km/h, so the answer is B. 5km/h.
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A person weighing 785 newtons on the surface of Earth would weigh 298 newtons on the surface of Mars.
rodikova [14]

Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²

Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

m = 785/9.8

m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

Thus the gravitational field strength on the surface of Mars = 3.72 m/s²

3 0
3 years ago
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Answer:

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Explanation:

5 0
3 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Lelechka [254]

Answer:

a. 16 s b. -1.866 kJ

Explanation:

a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.

We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².

Since the turntable stops at ω = 0, the time it takes to stop is gotten from

ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.

So it takes the turntable 16 s to stop.

b. The workdone by the turntable to stop W equals its rotational kinetic energy change.

So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ

3 0
3 years ago
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