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stellarik [79]
3 years ago
10

An object is moving at a constant velocity. All but one of the following statements could be true. Which one cannot be true?

Physics
1 answer:
likoan [24]3 years ago
6 0

Answer:

b.A single force acts on the object.

Explanation:

Newton's first law establishes that an object at rest remains at rest and an object in motion will have constant velocity if the result of the forces acting on them is zero

An object acceleration can be defined as the variation of velocity in an interval of time:                            

a = \frac{v_{f}-v_{i}}{t_{f}-t_{i}}   (1)  

Where v_{f} is the final velocity, v_{i} is the initial velocity,  t_{f} is the final time and t_{i} is the initial time.

If the velocity is constant there is no variation of velocity, therefore, the acceleration will be 0.

Newton's second law can be used To show that no forces act on the object:

R = ma (2)

Where R is the resultant of all the forces, m is the mass and a is the acceleration.

Since there is no acceleration for an object with constant velocity, equation (2) can be expressed as:

R = m(0)

R = 0  

So it is correct to say that an object with constant velocity has no forces acting on it.

In the other hand in the case of several forces acting on the object, The net force (R) can be zero if they have the same magnitude and opposite direction

So (a), (c) and (d) are correct because of what was previously discussed.

<u>Note that b cannot be true since it violates Newton's first law, since the object is moving at a constant velocity. </u>

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A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown
nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

v_x = 15.0 m/s

and with an initial vertical velocity of

v_{y0} = -20.0 m/s

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

y(t) = h + v_{0y} t + \frac{1}{2}gt^2 (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

v_y = -20 m/s

So the vertical position of the balloon after a time t is

y(t) = h + v_y t

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

v_x = 15.0 m/s

So the horizontal distance travelled in t = 6.00 s is

d_x = v_x t = (15.0 m/s)(6.00 s)=90 m

Considering also that the vertical height of the balloon after t=6.00 s is

d_y = 176.6 m

The distance between the balloon and the rock can be found by using Pythagorean theorem:

d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer at rest in the basket is

v_y (t) = gt

Substituting time t=6.00 s, we find

v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer on the ground is now given by

v_y (t) = v_{0y} + gt

Substituting time t=6.00 s, we find

v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s

6 0
3 years ago
a 0.05-kg starts from rest at a height of 0.95m. assuming no friction, what is the kinetic energy of the car when it reaches the
Alex73 [517]
Fhubugukhkbjhvlfipy0iy ft
3 0
3 years ago
The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c
Alex Ar [27]

Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

speed = 11 m/s

to find out

point due west from the current position

solution

we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west

so it become like triangle with 3.3 point down and the hypotenuse is 11

so by triangle

hypotenuse ×cos(angle) = adjacent side

11 ×cos(angle) = 3.3

cos(angle) = 0.3

angle = 72.54 degree west of south

3 0
3 years ago
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What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?
LenaWriter [7]

Answer:

Hello,~There!

What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?

<h2><u>33% According to the Graph</u></h2>

Hope this helps!

6 0
3 years ago
If a dog walks north for 10 meters and then east for 10 meters, what is the direction of its displacement?
Katyanochek1 [597]

The direction of its displacement wil be

c.northeast

In fact, the dog walks north for 10 meters and east for another 10 meters. The path of the dog can be represented with two vectors, A pointing north (of magnitude 10 meters) and B pointing east (of magnitude 10 meters). The direction of the resultant vector (due to east) will be given by

tan \theta =\frac{A}{B}=\frac{10}{10}=1

\theta=tan^{-1} (1)=45^{\circ}

and the direction will be north-east.

5 0
3 years ago
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