Categories

## If a force of 20 N is applied to an object with a mass of 3 kg, the object will accelerate at m/s2

If a force of 20 N is applied to an object with a mass of 3 kg, the object will accelerate at m/s2

Categories

## A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far has the boat traveled? Round your answer to the nearest whole number.

The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies.

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface.

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h.

So the time for the pencil and eraser to touch is T=1.3449 hours.

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

Categories

## A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)?

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)?

Categories

## A car initially moving at 9 m/s begins accelerating at its maximum acceleration of 4 m/s2. if the car maintains this acceleration, how much time will it take to cover 200 m?

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

Categories

## If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, how far would it have fallen (in m) after 2 s? (g = 10 m/s2) 20.0

The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies.

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface.

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h.

So the time for the pencil and eraser to touch is T=1.3449 hours.

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

Categories

## A ball of mass 0.5 kg is released from rest at a height of 30 m. How fast is it going when it hits the ground? Acceleration due to gravity is g = 9.8 m/s²

1) Water has very high specific heat.

2) It expands when it freezes

3) ability to dissolve in ionic substances

4) Water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid.

Explanation:

1) Water molecules due to their high specific heat , undergo relatively less increase in temperature after absorbing heat. So they prevent their adjoining area to experience sharp increase in temperature.

2) Water is an example which expands on freezing. In it , large spaces are left between molecules when it cools. It happens because hydrogen bonds prevent any two molecules to come very close thereby  creating gaps. This gap is filled up when ice melts , which results in overall reduction in volume of water.

3) Uneven distribution of charge results in water molecules becoming polar in nature . A polar substance  can be dissolved in polar solvents only.

4) It is due to property of surface tension. Property  of surface tension results due to molecules of water near the surface experiencing a  net downward  force due to hydrogen bonds . It is due to fewer hydrogen bonds made by them.

Categories

## A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far is the boat traveled?

A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far is the boat traveled?

Categories

## A motorist drives along a straight road at a constant speed of 15.0 m/s. Just as she passes a parked motorcycle police offi cer, the offi cer starts to accelerate at 2.00 m/s2 to overtake her. Assuming that the offi cer maintains this acceleration, (a) determine the time interval required for the police offi cer to reach the motorist. Find (b) the speed and (c) the total displacement of the offi cer as he overtakes the motorist.

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

Categories

## An object with an initial velocity of 3.0 m/s has a constant acceleration of 2.0 m/s2. When its speed is 19.0 m/s, how far has it traveled?

The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies.

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface.

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h.

So the time for the pencil and eraser to touch is T=1.3449 hours.

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

Categories

## You have designed a rocket to be used to sample the local atmosphere for pollution. It is fired vertically with a constant upward acceleration of 17 m/s2. After 30 s, the engine shuts off and the rocket continues rising (in freefall) for a while. (Neglect any effects due to air resistance.) The rocket eventually stops rising and then falls back to the ground. You want to get a sample of air that is 23 km above the ground.

The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies.

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface.

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h.

So the time for the pencil and eraser to touch is T=1.3449 hours.

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

Categories