All categories

1 year ago

What is the Physics Primer?

Physics

2 answers:

Elza [17]1 year ago

7 0

Answer:

A. a set of mathematically topics that are relevant to introductory physics.

Explanation:

The physics primer is not defined as the online comprehensive mathematics textbooks. It is the set of topics of mathematics which gives students trouble and remember.

Therefore, it is defined as the process of physics problem solving. So, mathematically skills are covered in physics course as a primer related success.

Therefore, it is a set of topics of mathematics that are relevent to introductory physics.

Dimas [21]1 year ago

5 0

Answer:

A. A set of mathematics topics that are relevant to introductory physics courses.

Explanation:

Physics Primer consists of tutorials, videos, and answers to specific questions, the main idea is to build a solid structure of math concepts for introductory physics. Another great thing about Physics Prime is that it helps students to visualize problems and solve them from different perspectives. The concepts built throughout the course can positively impact the result of the student especially if the course is taken before the physics introductory course.

You might be interested in

A person who climbs up something (e.g., a hill, a ladder, the stairs) from the ground gains potential energy. a person's weight

zhannawk [14.2K]

We are given the following values:

weight w = 240 lb = 1,067.52 N

energy E = 3,000 J

The formula for potential energy is:

E = w h

where h is the height the person has to climb, therefore:

h = 3000 / 1067.52

h = 2.81 m

Therefore he has to climb 2.81 meters

3 0

2 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the

sergeinik [125]

Answer:The weight of the door creates a CCW torque given by Tccw = 145 N*3.13 m / 2 You need a CW torque that's equal to that Tcw = F*2.5 m*sin20

4 0

2 years ago

Read 2 more answers

A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w

slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0

2 years ago

4. While broadcasting a football game, the announcer exclaimed, "I can't believe it. Carl James just scored a touchdown. That's

Zielflug [23.3K]

D. Carl doesnt score touchdowns very often.

7 0

2 years ago

Read 2 more answers

You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w

Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg mass of the box

d= 5.4m length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof and the y-axis in the direction perpendicular to it.

W: Weight of the box : In vertical direction

N : Normal force : perpendicular to the direction the roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight of the box

W= m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86 = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34 m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:

d:displacement = 5.4 m

v₀: initial speed

vf: final speed = 0

a : acceleration of the box = -1.34 m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

$v_{o} =\sqrt{14.472}$

v₀ = 3.8 m/s

7 0

2 years ago