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blsea [12.9K]
3 years ago
12

An object starts at rest. Its acceleration over 30 seconds is shown in the graph below:

Physics
1 answer:
ddd [48]3 years ago
6 0

Answer:

The instantaneous speed of the object after the first five seconds is 12.5 m/s.

(C) is correct option.

Explanation:

Given that,

An object starts at rest. Its acceleration over 30 seconds.

We need to calculate the instantaneous speed of the object after the first five seconds

We know that,

Area under the acceleration -time graph gives speed.

According to figure,

speed = area\ of\ tringle

speed=\dfrac{1}{2}\times b\times h

speed =\dfrac{1}{2}\times5\times5

speed-12.5\ m/s

Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.

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Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

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3 years ago
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3 years ago
How much pressure is applied to the ground
statuscvo [17]

Answer:

Pressure applied by the man= 285103.125 Pa  or 41.35 lb/in^{2}

Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e.  Pressure=\frac{Force}{Area}

Now, Force= mg

where, m = mass of the body(man) = 93 kg

g = acceleration due to gravity of Earth = 9.81 m/{s^{2}}

Area covered is equal to the area of both stilts(a man generally stands on two feet)

therefore Area=2(0.04)^{2} m^{2}

and putting in the values, we get,

Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}

Now we need to convert to our required units:

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Using the above relations we get,

Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}

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3 years ago
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A block of mass 9.2kg rests on a slope an angle of 26.9∘ relative to the horizontal. What is the size of the contact force norma
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Answer:

80.4 N

Explanation:

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We are only interested in the forces that act on the block along the direction perpendicular to the slope. Along this direction, we have two forces acting on the block:

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And by substituting the numbers into the equation, we find the size of the contact force normal to the slope:

N=(9.2 kg)(9.8 m/s^2)(cos 26.9^{\circ})=80.4 N

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3 years ago
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