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3 years ago

8

Which describes acceleration involving only a change in direction?

1 answer:

Viefleur [7K]3 years ago

6 0

If we consider any system moving with u<span>niform circular motion we can notice that the MAGNITUDE of the accelaration remains constant. However, there is a change in the direction of the acceleration at every instant of time .

As the object moves through the circle the acceleration changes its direction always pointing to the center of the circle.</span>

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Need help with this one question!

Lelu [443]

Answer:

Actual Mechanical Advantage

Explanation:

AMA=mg/F(applied)

8 0

2 years ago

Read 2 more answers

Which is a better conductor, metal or non metal

sweet [91]

Metal would be a better conductor cause electricity can travel easier through it

7 0

2 years ago

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

6 0

2 years ago

A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.

iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2$

Converting to m/s²

$a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

$\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g$

Hence, acceleration of the car is -0.64525g

8 0

3 years ago

Bad White [126]

Answer:

12.5 km

Explanation:

Since 30.0 minutes is half of an hour and the average speed is 25.0 km/h we can multiply the time by the speed to find the distance travelled.

25.0km/h x 0.5 hours = 12.5 km

4 0

2 years ago