Category Archives: Physics

Grantham PHY220 Week 2 Assignment Problem 1

A ship has a top speed of 3 m/s in calm water. The current of the ocean tends to push the boat at 2 m/s on a bearing of due South. What will be the net velocity of the ship if the captain points his ship on a bearing of 55° North of West and applies full power?

Solution:

Week 2 Problem 1

$R_x=-3\:cos\:55^{\circ }=-1.720729309\:m/s$

$R_y=3\:sin\:55^{\circ }-2=0.4574561329\:\:m/s$

The x component of the resultant is negative and the y component is positive, thus the resultant is located at the second quadrant.

$R=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}=\sqrt{\left(-1.720729309\:m/s\right)^2+\left(0.4574561329\right)^2}=1.78\:m/s$

$\theta =tan^{-1}\left(\frac{R_y}{R_x}\right)=tan^{-1}\left(\frac{0.4574461329}{-1.720129309}\right)=-14.9^{\circ}$

Therefore, the magnitude of the net velocity of the ship is 1.78 m/s, and is going 14.9 degrees North of West

Grantham PHY220 Week 2 Assignment

Problems:

A ship has a top speed of 3 m/s in calm water. The current of the ocean tends to push the boat at 2 m/s on a bearing of due South. What will be the net velocity of the ship if the captain points his ship on a bearing of 55° North of West and applies full power? [Solution]
An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y-axis and the total distance to the plane from you. Choose the x-axis east, y-axis north, and z-axis up. [Solution]
A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s? [Solution]
A not so brilliant physics student wants to jump from a 3rd-floor apartment window to the swimming pool below. The problem is the base of the apartment is 8.00 meters from the pool’s edge. If the window is 20.0 meters high, how fast does the student have to be running horizontally to make it to the pool’s edge? [Solution]
If a 1500 kg car stopped from an in 5.6 seconds with an applied force of 5000 N, how fast was it initially traveling?[Solution]
If the acceleration due to gravity on the Moon is 1/6 that what is on the Earth, what would a 100 kg man weight on the Moon? If a person tried to simulate this gravity in an elevator, how fast would it have to accelerate and in which direction? [Solution]
A 7.93 kg box is pulled along a horizontal surface by a force F_p of 84.0 N applied at a 47.0° angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box? [Solution]
If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road? [Solution]

Grantham PHY220 Week 1 Assignment Problem 8

Problem 8

A stone is dropped from the roof of a high building. A second stone is dropped 1.25 s later. How long does it take for the stones to be 25.0 meters apart?

Solution:

Let t be the amount of time after the first stone is dropped. The distance from traveled by the first stone is

$y_1=\frac{1}{2}gt^2$

The distance traveled by the second stone is

$y_2=\frac{1}{2}g\left(t-1.25\right)^2$

The difference between the two stones is 25.0 m after time

$y_1-y_2=25.0$

$\frac{1}{2}\left(9.80\:m/s^2\right)t^2-\frac{1}{2}\left(9.80\:m/s^2\right)\left(t-1.25\right)^2=25$

$4.9t^2-4.9\left(t^2-2.5t+1.5625\right)=25.0$

$4.9t^2-4.9t^2+12.25t-7.65625=25.0$

$12.25t=25.0+7.65625$

$12.25t=32.65625$

$t=2.67\:s$

Grantham PHY220 Week 1 Assignment Problem 7

Problem 7

Explain a possible situation where you start with a positive velocity that decreases to a negative increasing velocity while there is a constant negative acceleration.

Solution:

An example of this situation is a free-fall. If an object is thrown upward, the initial velocity is positive. Then the velocity decreases until the object thrown will reach its maximum height and then it goes back with a negative increasing velocity. In this entire flight, the acceleration is a constant negative–the acceleration due to the Earth’s gravity.

Grantham PHY220 Week 1 Assignment Problem 6

Problem 6

A sports car moving at constant speed travels 150 m in 4.00 s. If it then brakes and comes to a stop while decelerating at a rate of 6.0 m/s² , how long does it take to stop?

Solution:

The velocity before applying the brake is

$v=\frac{150\:m}{4.00\:s}=37.5\:m/s$

Solve for t until the velocity becomes zero

$v=v_0+at$

$t=\frac{v-v_0}{a}=\frac{0-37.5\:m/s}{-6.0\:m/s^2}=6.25\:s$

Grantham PHY220 Week 1 Assignment Problem 5

Problem 5

A car travels 120 meters in one direction in 20 seconds. Then the car returns ¾ of the way back in 10 seconds. Calculate the average speed of the car for the first part of the trip. Find the average velocity of the car.

Solution:

The average speed of the car for the first part of the trip is

$average\:speed=\frac{x}{t}=\frac{120\:m}{20\:s}=6\:m/s\:$

The average velocity of the car is

$\overline{v}=\frac{\Delta x}{\Delta t}=\frac{120-\frac{3}{4}\left(120\right)\:m}{20+10\:s}=\frac{30\:m}{30\:s}=1\:m/s$

Grantham PHY220 Week 1 Assignment Problem 4

Problem 4

What is larger 4000 L or $4\times 10^5\:cm^3$ (and you must show your reasoning.)

Solution:

There are $1000\:or\:1\times 10^3\:cm^3$ in 1 L. So, in $4\times 10^5\:cm^3$ there are

$4\times 10^5\:cm^3\times \left(\frac{1\:L}{1\times 10^3\:cm^3}\right)=400\:L$

Therefore, the 4000 L is larger than $4\times 10^5\:cm^3$ .

Grantham PHY220 Week 1 Assignment Problem 3

Problem 3

What is the surface area of a sphere of diameter $2.4\times 10^2$ cm?

Solution:

$SA=4\pi r^2=4\pi \left(\frac{2.4\times 10^2\:cm}{2}\right)^2=1.8\times 10^5\:cm^2$

Grantham PHY220 Week 1 Assignment Problem 2

Problem 2

Multiply $1.783\times 10^{-2}\cdot 4.4\times 10^{-3}$ , taking into account significant figures.

Solution:

$1.783\times 10^2\cdot 4.4\times 10^{-3}=0.78\:or\:7.8\times 10^{-1}$

Grantham PHY220 Week 1 Assignment Problem 1

Problem 1

One lightyear is defined to be the distance light can travel in one year. What is this distance in meters? How long does it take for light to get to the Moon?

Solution:

$1\:lightyear=9.461\times 10^{15}\:m$

The approximate distance from Earth to the moon is 384,400 kilometers. The time it takes for the light to reach the moon is

$t=\frac{384400\:km\:\left(\frac{1000\:m}{1.00\:km}\right)}{299\:792\:458\:\frac{m}{s}}=1.28\:s$

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Category Archives: Physics

Grantham PHY220 Week 2 Assignment Problem 1

A ship has a top speed of 3 m/s in calm water. The current of the ocean tends to push the boat at 2 m/s on a bearing of due South. What will be the net velocity of the ship if the captain points his ship on a bearing of 55° North of West and applies full power?