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3 years ago

The unit meters corresponds to what variable

Physics

1 answer:

Sati [7]3 years ago

7 0

Do you have any options? My guess would be distance but I could be wrong.

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A 0.060-kg tennis ball, moving with a speed of 5.50m/s , has a head-on collision with a 0.090-kg ball initially moving in the sa

Yuliya22 [10]

Answer:

Explanation:

For elastic collision , the formula for velocity of .06 kg is

v₁ = (m₁-m₂)/(m₁+m₂) u₁ + 2m₁m₂/(m₁+m₂) u₂

=( .06-.09 / .06+.09 ) 5.5 + (2 x .06 x .09 / .06+.09 ) 3.4

=( -.03 / .15) x 5.5 + (2 x .0054 / .15) x 3.4

= -1.1 +.2448

= - 0 .8552 m/s

Its direction will be - opposite direction

the formula for velocity of .09kg is

v₂ = (m₂-m₁)/(m₁+m₂) u₂ + 2m₁m₂/(m₁+m₂) u₁

= ( .09-.06 / .06+.09 ) 3.4 + (2 x .06 x .09 / .06+.09 ) 5.5

= .68 +.396

= 1.076 m / s

Its direction will be in the same direction .

4 0

3 years ago

If the coefficient of static friction between your friend and the car seat is 0.500 and you keep driving at a constant speed of

Rasek [7]

Answer:

90m

Explanation:

Let g = 9.8 m/s2. The friction is the product of normal force and its coefficient, and normal force is equal to gravity

$F_f = \mu N = \mu mg$

The acceleration caused by friction, according to Newton's 2nd law:

$a_f = F_f / m = \mu g = 0.5 * 9.8 = 4.9 m/s^2$

For the friend to slide over, then the centripetal acceleration must be equal to friction acceleration.

Since you are driving at a constant speed of 21 m/s, then your maximum radius of curvature can be calculated using the following formula:

$a_c = a_f = v^2/r = 4.9$

$r = v^2/4.9 = 21^2/4.9 = 90 m$

7 0

3 years ago

8. The legs of a young man are each 0.650 meters long. What is his maximum walking speed?

Norma-Jean [14]

Answer:

2.52 m/s

Explanation:

When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.

Sum of the forces in the radial direction:

∑F = ma

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 0.650 m:

v = √(9.8 m/s² × 0.650 m)

v = 2.52 m/s

8 0

3 years ago

Why are Watson’s experiments considered controversial?

iris [78.8K]

D, it is considered unethical today

4 0

3 years ago

Read 2 more answers

The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad

Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

$A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})$

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

$A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )$

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

T = 2*p / w

T = 2*p / (2*p / 92.8)

Hence, T = 92.8 min

7 0

3 years ago