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2 years ago

If you separate vector B into its components. How many components will it have? Those components will be called?

Physics

1 answer:

Inessa [10]2 years ago

3 0

The vector B will have two components and those components will be called resultant vectors.

<h3>What is a component vector?</h3>

A component vector is a unit vector that represents a given vector in a particular direction.

A vector can be represented in x - direction and y - direction.

x - component of the vector = Bcosθ
y - component of the vector = Bsinθ

Thus, the vector B will have two components and those components will be called resultant vectors.

Learn more about component vectors here: brainly.com/question/13416288

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An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is t

xeze [42]

Answer:

The magnitude of the tension in he string is equal to the magnitude of the weight of the object.

Explanation:

According to the Newton's 1st law, An object will remain at rest or in uniform motion in a straight line unless acted upon by an unbalanced force.

In here, the elevator is moving with a constant speed. So the object must have the equal constant speed. Which means, it has a uniform motion. According to Newton's 1st law, the total unbalanced force on the object must be zero . As we know, there are only two forces are on the object and they are,

The tension in string(T) , The weight of the object(W) .

∴ F = 0

T - W = 0

So to balanced those forces, the magnitude of the tension in the string must be equal to the magnitude of the weight of the object.

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3 years ago

What are at least 3 examples of sublimation?

Step2247 [10]

Answer:

dry ice, air fresheners, polar evaporation, arsenic treatment

5 0

3 years ago

Read 2 more answers

A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe

joja [24]

Answer:

The Doppler Effect is given by the following relation;

$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

Explanation:

7 0

3 years ago

A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it

Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given: </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.

<u>Solution </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next

Ef=Ei+W (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form

(Kf + mc^2) = (KJ+ mc^2) (2)

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by

W = FΔr (4)

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut

W = F Δr= (190N)(6 m) = 1140 J

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W' (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate

1/2mvf'^2=1/2mvi'^2+W'

vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0

3 years ago

A bus and a bicycle have a head-on collision. Compare the force of impact between the bus and the bicycle; compare the accelerat

kati45 [8]

The force of impact is same for both bus and the bicycle. The acceleration of bicycle will be greater than the acceleration of bus.

<u>Explanation:</u>

The interaction that occurs between two objects refers to collision. this makes the two objects to come in contact with each other. The third law of Newton states that, when there occurs a collision between two objects, then the force that is applied on each object will be same. But, the direct in which the force is impacted will be in opposite direction.

The magnitude of the forces will be equal but the direction will not be same. The collision results in gaining the momentum by one object and losing momentum by another. The acceleration is mainly associated with the mass of the object. When the object has smaller mass, it will be accelerated more. In the given example, as bus is heavier than bicycle, the bicycle will have greater acceleration than the bus.

4 0

3 years ago