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padilas [110]
2 years ago
8

an 2-kg object is moving horizontally with a speed of 4m/s. how much net force os required to keep the object movong with the sa

me speed and in the same direction

Physics
1 answer:
Zepler [3.9K]2 years ago
7 0
In order to keep the object moving in the same direction and with same speed, its initial and final momentum should be zero.
∴ Net force on the object should be zero.
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An ambulance moving at 42 m/s sounds its siren whose frequency is 450 hz. a car is moving in the same direction as the ambulance
Korvikt [17]
(a) Since the ambulance and the car are moving one relative to each other, we have to use the general formula of the Doppler effect, which gives us the shift of the frequency of the siren as heard by an observer in the car:
f'=( \frac{v+v_o}{v+v_s} )f
where
f' is the apparent frequency as heard by the observer in the car
v is the velocity of the wave 
v_o is the velocity of the observer (positive if it is moving towards the source, negative if it is moving away)
v_s is the velocity of the source (positive if the source is moving away from the observer, negative if is is moving towards it)
f is the real frequency of the sound

In the first part of the problem:
v=343 m/s (speed of the sound wave)
v_o =-25 m/s (the car is moving away from the ambulance)
v_s = -42 m/s (the ambulance is moving towards the car)
f=450 Hz (original frequency of the sound)

If we plug the numbers into the formula, we find
f'=( \frac{343 m/s-25 m/s}{343 m/s-42 m/s} )(450 Hz)=475 Hz

b) This time, the ambulance passes the car, so the ambulance is now moving away from the car; this means that v_s must be positive:
v_s=+42 m/s
Moreover, the car is now moving towards the ambulance, so we should reverse also the sign of v_o:
v_o=+25 m/s
All the other data do not change, so if we use the same formula as before, we find
f'=( \frac{343 m/s+25 m/s}{343 m/s+42 m/s} )(450 Hz)=430 Hz
8 0
2 years ago
The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at
elena-14-01-66 [18.8K]

The position function x(t) of a particle moving along an x axis is x=4.00 - 6.00t^2

a) The point at which particle stop, it's velocity = 0 m/s

  So dx/dt = 0

        0 = 0- 12t = -12t

  So when time t= 0, velocity = 0 m/s

    So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

 x=4.00 - 6.00*0^2 = 4m

SO at x = 4m the particle is (momentarily) stop

c) We have x=4.00 - 6.00t^2

   At origin x = 0

  Substituting

         0 = 4.00 - 6.00t^2\\ \\ t^2 = \frac{2}{3}

         t = 0.816 seconds or t = - 0.816 seconds

So when  t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.

5 0
2 years ago
A projectile of mass m is fired horizontally with an initial speed of v0​ from a height of h above a flat, desert surface. Negle
Grace [21]

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

5 0
2 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
QveST [7]

Answer:

Please see attachment

Explanation:

Please see attachment

8 0
2 years ago
Which of the following best describes an action-reaction pair? A. The Moon Pulls on Earth, and Earth pulls back on the moon. B.
Papessa [141]
An action-reaction pair would be a pair in which one of the elements exerts a force on the other element (action), and then the other element would respond to this force by exerting another force in the opposite direction (reaction).

From the given choices, we will see that:
For choice A, the moon exerts a force on the earth by pulling it (action) and the earth responds to this force by pulling the moon (reaction in opposite direction of the action).

Therefore, the correct choice would be: 
A. <span>The Moon Pulls on Earth, and Earth pulls back on the moon.</span>
4 0
2 years ago
Read 2 more answers
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