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3 years ago

What is the meaning of Constant Acceleration?

1 answer:

4 0

Any change in speed or direction of motion is acceleration.

Constant acceleration can mean ...

-- speeding up at a constant rate . . . gaining the same amount
of speed each second.

-- slowing down at a constant rate . . . losing the same amount
of speed each second.

-- changing direction at a constant rate . . . for example, going
around a circular path at a constant speed.

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A motor is used to produce

Tpy6a [65]

Your answer would be 4.0 hz.

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3 years ago

A vector is given by R 2i + j+3k. Find (a) magnitudes of the x, y, and z components; (b) the ma nitude of R; and (c) the angles

Darya [45]

Answer:

given,

R = 2i + j+3k

a) magnitude in x = 2

y = 1

z = 3

b) magnitude of R

$R = \sqrt{2^2+1^2+3^2}$

R = 3.74 units

c) angle between the R and the x, y, and z axes.

$cos \theta_x=\dfrac{2}{3.74}$

θ x = 57.72°

$cos \theta_y=\dfrac{1}{3.74}$

θ y = 74.51°

$cos \theta_z=\dfrac{3}{3.74}$

θ z = 36.68°

4 0

3 years ago

Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler

ehidna [41]

A) Acceleration in part A: $6.1 m/s^2$

B) Acceleration in part B: $2.7 m/s^2$

C) Acceleration in part C: $1.5 m/s^2$

Explanation:

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

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

where

v is the final velocity

u is the initial velocity

$\Delta t$ is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

$\Delta t = 0.9 - 0 = 0.9 s$

So the acceleration is

$a=\frac{5.5-0}{0.9}=6.1 m/s^2$

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

$\Delta t = 2.4 - 0.9 = 1.5 s$ is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

$a=\frac{9.5-5.5}{1.5}=2.7 m/s^2$

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

$\Delta t = 3.4 - 2.4 = 1 s$ is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

$a=\frac{11-9.5}{1}=1.5 m/s^2$

Learn more about acceleration:

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#LearnwithBrainly

3 0

3 years ago

A plastic rod and a piece of cloth are both uncharged. A student rubs the plastic rod with the cloth. The plastic rod becomes ne

Molodets [167]

Answer: a sock and a Pepe when the Pepe is rubbed by the sock the energy is enormous

Explanation:

7 0

3 years ago

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr

IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

$P=\frac{V^2}{R}$

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is $P=5.0 W$ , and the resistance of the resistor is $R=15 k\Omega = 15000 \Omega$ , then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V: