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Andrej [43]
4 years ago
13

Which statement correctly describes the differences between positive and negative acceleration? Positive acceleration describes

a change in speed; negative acceleration describes no change in speed. Positive acceleration describes no change in speed; negative acceleration describes a change in speed. Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed. Positive acceleration describes a decrease in speed; negative acceleration describes an increase in speed.
Physics
2 answers:
salantis [7]4 years ago
7 0

Answer:

Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed.

Explanation:

Acceleration is defined as:

a=\frac{v-u}{\Delta t}

where

v is the final velocity

u is the initial velocity

\Delta t is the time taken for the change in velocity to occur

The value of \Delta t is always positive, so the sign of the acceleration depends on the numerator. More specifically, we have two situations:

- if v > u, then it means that the final velocity is greater than the initial velocity, so the acceleration is positive and the speed is increasing

- if v < u, then it means that the final velocity is smaller than the initial velocity, so the acceleration is negative and the speed is decreasing

qwelly [4]4 years ago
4 0

The correct answer is - Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed.

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A wave pulse travels along a string at a speed of 200 cm/s. What will be the speed if:
34kurt

Answer:

a) v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s

The velocity increase by a factor of \sqrt{2}

b) v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s

The velocity decrease by a factor of 2.

c) v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s

The velocity increase by a factor of 2

d) v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s

The velocity not changes.

Explanation:

For this case we know that the velocity is v = 200 cm/s = 2m/s

v_f represent the final velocity after the changes specified,

Part a

The formula for the speed of a wave in a string is given by:

v = \sqrt{\frac{T}{\rho}}

And the linear density is defined as:

\rho = \frac{m}{L}

And if we replace this we got:

v = \sqrt{\frac{TL}{m}}

If the tension mass is doubled we have this:

v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s

The velocity increase by a factor of \sqrt{2}

Part b

If we mass is quadrupled we have this:

v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s

The velocity decrease by a factor of 2.

Part c

If the length is quadrupled we have this:

v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s

The velocity increase by a factor of 2

Part d

For this case we know that the mass and the length are both quadrupled and we got:

v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s

The velocity not changes.

7 0
3 years ago
The electric field is strongest where equipotential curves are:_______.A) closest togetherB) farthest apartC) most nearly straig
cluponka [151]

Answer:

A closest

Explanation:

This is because the electric field will be strongest or largestwhen the equipotential curves are closest together

We know that the field is

E= V/d

Where is distance and we see that d being the denominator will only make E bigger if it becomes smaller that is the curves closest

3 0
3 years ago
a baby carriage is sitting at the top of a hill that is 21m high. the carriage with the baby weighs 25kg. calculate the speed th
Ulleksa [173]

Answer:

Velocity = 20.3 [m/s]

Explanation:

This is a typical problem of energy conservation, where potential energy is converted to kinetic energy. We must first find the potential energy. In this way, we will choose as a reference point or point where the potential energy is zero when the carrriage rolls down 21 [m] from the top of the hill.

E_{p} =m*g*h\\ where:\\m = mass = 25[kg]\\g = gravity = 9.81 [m/s^2]\\h = elevation = 21 [m]\\E_{p} =potential energy [J]\\E_{p} =25*9.81*21=5150[J]

Now this will be the same energy transformed into kinetic energy, therefore:

E_{p}=E_{k} = 5150[J]\\E_{k} =0.5*m*v^{2} \\where:\\v=velocity [m/s]\\v=\sqrt{\frac{E_{k}}{0.5*25} } \\v=20.3[m/s]

3 0
3 years ago
Does the vertical or horizontal position affect an objects potential energy?
son4ous [18]
Yes. If an object would be vertical, it would most likely have potential energy due to gravity. 
6 0
3 years ago
The bar ab has a velocity at point a of va = [20,9] m/s and an angular velocity w = 23 rad/s. If the distance between a and b is
DiKsa [7]

speed of point A is given as

v_a = 20.9 m/s

v_a = r*\omega

here r = distance from the axis

so here we have

20.9 = r*23

r  = 0.91 m

now the distance of point A and B is 0.71 m

while the distance of axis from point A is 0.91 m

so distance of axis from point B will be given as

r = 0.91 + 0.71 = 1.62 m

now for the speed of point b is given as

v = r \omega

v = 1.62 * 23

v = 37.26 m/s

so end point B will move with speed 37.26 m/s

8 0
3 years ago
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