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3 years ago

It takes a car 2 minute(s) to go from rest to 25 m/s east. What is the acceleration

Physics

1 answer:

yKpoI14uk [10]3 years ago

5 0

Explanation:

we first convert 2 minutes to seconds,which will give us 120s (2×60).

v=u +at where (v=final velocity,u is the initial velocity,a is the acceleration of the car and t the time taken)

V=25 m/s

u=0 m/s since the car started from rest 

t=120s

substituting the values we get 

25=0+a (120)

25=120a (diving both sides by 120 we get)

a=0.208m/s^2

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A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co

ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block. There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0

3 years ago

Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,

jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

Therefore: $E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}$

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$

7 0

3 years ago

A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in

vovangra [49]

Answer:

L = Pt/M

Explanation:

Power, P= Q/t = mL/t

we know that, (Q=m×l)

Now ⇒l= Pt/M

Thus l= Pt/M

4 0

3 years ago

An iron nail goes through a chemical change when iron in the nail combines with oxygen in the air to form iron oxide. This compo

Mazyrski [523]

It is like that, except most nails are steel or stainless steel, slowing to rusting process to about 5 years.

6 0

3 years ago

Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m), 2.0 kg at (2.0 m, 0.0 m), and 4.0 kg at (2.0 m, 1.0 m).

melamori03 [73]

Refer to the diagram shown below.

The given data is

mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)

Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.

Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m

8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m

Answer: (1.5, 0.5) m

6 0

3 years ago