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Anna007 [38]
3 years ago
5

Compare and Contrast the following pairs of terms.

Physics
1 answer:
Salsk061 [2.6K]3 years ago
4 0
1- Speed is the modulus or magnitude (scalar) of the velocity (vector)
2- Motion is the length of the trajectory followed while displacement is the final position - the initial position, xf-x0.
3-  Here it's referring to the linear momentum. An object moving in a straight line has a momentum p=m*v. Where m is the mass of the object and v is the velocity.
4- Acceleration is the rate of change of velocity with respect to time. a=dv/dt.
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Answer:

leaf and a balloon is the correct answer

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What type of energy is formed when an object encounters friction?​
kiruha [24]

Answer:

thermal energy

Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force. The forces acting on the package are gravity, the normal force, the force of friction, and the applied force.

Explanation:

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Dish-shaped reflectors are used to steer microwaves in order to establish communications links between nearby buildings. Those r
Ad libitum [116K]

A "screen" or even just a set of parallel bars are highly reflective to electromagnetic waves as long as the open spaces are small compared to the wavelengths.

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(I even worked on a microwave system in South America where huge grid dishes were used on a 90-mile link.)

5 0
2 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0
3 years ago
A 3 kg toy car with a speed of 7 m/s collides head-on with a 2 kg car traveling in the opposite direction with a speed of 5 m/s.
Mademuasel [1]

Answers:

kinetic energy lost = 86.4J

Explanation:

let Kf be the kinetic energy after the collision and Ki be the kinetic energy before the collision. let the 3kg car be 1 and 2kg car be 2.

Kf = K1(f) + K2(f)

Ki = K1(i) + k2(i)

loss in kinetic energy = Kf - Ki

                                 = 1/2(3)(2.20)^2 + 1/2(2)(2.20)^2 - 1/2(3)(7)^2 - 1/2(2)(-5)^2

                                 = 12.1 - 98.5

                                 = -86.4 J

therefore, the kinetic energy lost in the collision is 86.4 J.

6 0
3 years ago
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