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2 years ago

This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14

Physics

1 answer:

Harrizon [31]2 years ago

3 0

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

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A car traveling at 37m/s starts to decelerate steadily. It comes to a complete stop in 15 seconds. What is it’s acceleration

LuckyWell [14K]

Answer

The acceleration is

$a=-2.5ms^{-2}$ to the nearest tenth

Explanation

Since the car was travelling at $37ms^{-1}$ before it starts to decelerate, the initial velocity is

$u=37ms^{-1}$ .

The final velocity is $v=0ms^{-1}$ , because the car came to a stop.

The time taken is $t=15s$ .

Using the Newton's equation of linear motion,

$v=u +at$ , we find the acceleration by substituting the known values.

This implies that,

$0=37 +a(15)$

This gives us,

$0-37=15a$

$\Rightarrow -37=15a$

We divide both sides by 15 to get,

$a=-\frac{37}{15}ms^{-2}$

$a=-2.46667ms^{-2}$

6 0

3 years ago

You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion

Nina [5.8K]

Answer:

v (minimum speed) = 2.90 m/sec.

$\\ \\ maximum speed (v)= 6.57 m/sec.\\$

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?

Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

$m*g + T = m*v^2/R$

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

$v = \sqrt{g*R}$

$v = \sqrt{9.81*0.86}$

v (minimum speed) = 2.90 m/sec.

b) what is the maximum speed the rock can have so that the string does not break?

Here the force balance at bottom of circle is represented by the illustration:

$T = m*g + m*v^2/R$

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass m = 0.75 kg

∴

$45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\$

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because at the lowest point; the tension in string will be maximum.

4 0

2 years ago

If a machine has an efficiency of 50% and an input of 3 J, what is the output?

uranmaximum [27]

Answer:

150J

Explanation:

work output/work input=100%

so just make work output the subject

6 0

2 years ago

Anything that has mass and occupies space is defined as.

forsale [732]

Answer:

matter is the correct answer.

8 0

2 years ago

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains

OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

C = 771.35 J/kg°C

5 0

3 years ago