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Ostrovityanka [42]

3 years ago

12

As the mass of an object in uniform circular motion increases, what happens to the centripetal force required to keep it moving

at the same speed?

1 answer:

Dmitriy789 [7]3 years ago

7 0

Answer:

It increases, because the centripetal force is directly proportional to the mass of the rotating body.

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Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

ira [324]

Answer:

$\Delta L=15\,mm$

Explanation:

Given:

length of a steel-string, $L=1m$

area of the string, $A=0.5\,mm^2$

Young's modulus of the steel, $Y=2\times 10^{11} Pa$

force of tension on the string, $F=1500\,N$

We have the relation for change in length:

$\Delta L=\frac{F.L}{A.Y}$

$\Delta L=\frac{1500\times 1000}{0.5\times 10^{-6}\times 2\times 10^{11}}$

$\Delta L=0.015m$

$\Delta L=15\,mm$

6 0

3 years ago

A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?

Anarel [89]

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>:</em><em>)</em>

8 0

3 years ago

Read 2 more answers

What is a lenticular (S0) galaxy? A galaxy with a disk and central bulge like a spiral galaxy, but with no spiral arms A galaxy

RSB [31]

Answer:

A galaxy with a disk and central bulge like a spiral galaxy, but with no spiral arms

Explanation:

A Lenticular galaxy is a kind of galaxy intermediate between elliptical galaxy and a spiral galaxy in the Morphological classification system of galaxies. They have a central bulge or disc just like a Spiral galaxy but lacks the arms of spiral galaxy. If looked edge on they appear to be spiral and if looked face on they appear to be elliptical.

The absence of spiral arms can be attributed to the absence of star formation. They mainly consists of ageing stars.

6 0

4 years ago

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0

3 years ago

Name three ways a workout in the healthy heart zone benefits your health

babymother [125]

Peak. Intensity: Vigorous (85 to 100 percent of your max heart rate) Benefit: Increases performance speed.
Cardio. Intensity: Hard (70 to 84 percent of your max hr) Benefit: Builds cardiovascular fitness and muscle strength.
Fat Burn. Intensity: Moderate (50 to 69 percent of your max hr)

7 0

3 years ago