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garri49 [273]
2 years ago
8

The first step in a star formation is ______?

Physics
2 answers:
Mariana [72]2 years ago
8 0
First look at the star
AleksandrR [38]2 years ago
8 0
Initial collapse of an interstellar cloud
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What are some similarities between heat and temperature? 
dalvyx [7]
<span><u>Answer </u>
Heat is a form of energy while temperature is the measure or degree of coldness or hotness of a substance, place or something.
The similarities between heat and temperature are:
1. Heat causes the temperatures to raise. This means that it is the heat energy that causes temperatures to change.
2. Both heat and temperatures can be measured by specific units. Heat is measured in joules while temperature is measured in kelvin.  
</span>
7 0
3 years ago
Read 2 more answers
If you drop a ball off a cliff, it starts out a 0 m/s. After 1 s, it will be traveling at about 10 m/s. If air resistance is rem
Softa [21]

Answer: (C) The ball will accelerate to about 20 m/s.

Explanation: The ball is in free fall and no other but gravitational force acts on it. This means the ball will be subject to the gravitational acceleration, which is about 9.8 m/s^2. So after one additional second the ball will increase its velocity to (10+9.8) m/s = 19.8 m/s or approx. 20 m/s.

8 0
3 years ago
Read 2 more answers
A girl is sitting in a sled sliding horizontally along some snow (there is friction present).
prisoha [69]

Answer:

Ff = 159 [N]

Explanation:

Since in the problem give us two values of friction coefficient one static and the other dynamic, it is important to keep in mind that in order to solve this problem we must know the conditions in which the sled is located. We can read that the sled is in motion, in this way we use the dynamic coefficient of friction.

The friction force is calculated as the product of normal force, by the appropriate coefficient of friction value.

Ff = N * μ

Ff = 636 * 0.25

Ff = 159 [N]

6 0
3 years ago
A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a
grigory [225]

Answer:

a)  fr = 266.92 N,   fy = 1300 N,  b)    μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

             -w x - W x₂ + R y = 0         (1)

usemso trigonometry to find distances

            cos 60.08 = x / 7.5

            x = 7.5 cos 60.08

            x = 3.74 m

fireman

           cos 60.08 = x₂ / 4

           x2 = 4 cos 60

           x2 = 2 m

wall support

           sin 60.08 = y / 15

           y = 15 are 60.08

           y = 13 m

we substitute in equation 1

           R y = w x + W x2

            R = (w x + W x2) / y

            R = (500 3.74 +800 2) / 13

            R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

           R -fr = 0

           R = fr

           fr = 266.92 N

Y Axis  

           Fy - w-W = 0

           fy = 500 + 800

           fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

          cos 60.08 = x₃ / 9.00

          x₃ = 9 cos 60

          x₃ = 5.28 m

from equation 1

          R = (w x + W x₃) / y

          R = 500 3.74 + 800 5.28) / 13

          R = 468.769 N

we saw that

          fr = R = 468.769

The expression for the friction force is

          fr = μ N

in this case the normal is the ratio to pesos

        N = Fy

       N = 1300 N

        μ N = fr

        μ = fr / N

        μ = 468,769 / 1300

         μ = 0.36

7 0
3 years ago
Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity
Anna11 [10]

Answer:

the final velocity of the two blocks is v = \frac{mv_o}{m+M}

the distance that A slides relative to B is S = \frac{v_o^2M}{2 \mu g (M+m)}  

Explanation:

From the diagram below;

acceleration of A relative to B is : a = - ( \mu g  + \frac{ \mu mg}{M})

where

v = u + at

0 = v_o + ( - \mu g - \frac{\mu m g }{M})t

Making t the subject of the formula; we have:

t = \frac{v_o M}{(\mu g )(M+m)}

v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\

S = \frac{v_o^2M}{2 \mu g (M+m)}  which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\

v = \frac{mv_o}{m+M}

Thus, the final velocity of the two blocks is v = \frac{mv_o}{m+M}

4 0
2 years ago
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