Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placing it in orbit? What will happen if the horizontal speed is too small? What will happen if the speed is too large?

Answer:

The magnitude of the average force on the wall during the collision is 6 N.

Explanation:

Given;

mass of snowball, m = 120 g = 0.12 kg

velocity of the snowball, v = 7.5 m/s

duration of the collision between the snowball and the wall, t = 0.15 s

Magnitude of the average force can be calculated by applying Newton's second law of motion;

F = ma

where;

a is acceleration = v / t

a = 7.5 / 0.15

a = 50 m/s²

F = ma

F = 0.12 x 50

F = 6 N

Therefore, the magnitude of the average force on the wall during the collision is 6 N.

Explanation:

If Team A is on the left, B is on the right

if the force is constant, it means that the effort applied is equal.

So Team B is pulling 5000N to the right.

Answer:

1.01 W/m

Explanation:

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

From the above, we assumed that the pipe wall and the oil are in thermal equilibrium.

area of the pipe per unit length A = [tex]\pi r ^{2}[/tex] = [tex]7.069*10^{-4}[/tex] m^2/m

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130 = 1.01 W/m

The heat loss per unit length of tube should be considered as the 1.01 W/m.

Since

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

Now

area of the pipe per unit length A should be

= πr^2

 = 7.069*10^-4 m^2/m

Now

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130

= 1.01 W/m

hence, The heat loss per unit length of tube should be considered as the 1.01 W/m.

Learn more about heat here: https://brainly.com/question/15170783

As waves travel into the denser medium, they slow down and wavelength decreases.

Explanation:

The denser the medium the slower the waves (speed of light) travels.

◦•●◉✿When the body approaches the speed of light, the body's length appears to contract in the direction of travel, and its mass appears to increase from the point of view of a stationary observer. Only photons move to light velocity. They don´t have length.✿◉●•◦

Answer:

d. Both kinetic and sliding friction

Explanation:

Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.

Answer:

E = 85170 J (/ 85.2 kJ)

Explanation:

Take the latent heat of fusion of water be 334J / g.

From the equation E = ml,

E = energy required (unknown),

mass m = 255g,

latent heat of fusion l = 334J / g,

E = 255 x 334

E = 85170 J (/ 85.2 kJ)

Answer:

The wavelength is approximately 611 nm

Explanation:

We can use the formula for the condition of maximum of interference given by:

[tex]d\,sin(\theta)=m\,\lambda\\(0.000250\,\,m)\,\,sin(1.12^o)=8\,\lambda\\\lambda=\frac{1}{8} \,(0.000250\,\,m)\,\,sin(1.12^o)\\\lambda \approx 610.8\,\,nm[/tex]

We can also use the formula for the distance from the central maximum to the 5th minimum by first finding the tangent of the angle to that fifth minimum:

[tex]tan(\theta)=\frac{y}{D}\\ tan(\theta)=\frac{0.0333}{3.02} =0.011026[/tex]

and now using it in the general formula for minimum:

[tex]d\,sin(\theta)\approx d\,tan(\theta)=(m-\frac{1}{2} )\,\lambda\\\lambda\approx 0.00025\,(0.011026)/(4.5)\,\,m\\\lambda\approx 612.55\,\,nm[/tex]

Answer:

The correct answer is  [tex]6.1\times10^{-7}\:m[/tex]

Explanation:

The distance from the central maxima to 5th minimum is:

[tex]x_{5n}-x_{0} =3.33\:cm=0.033\:m[/tex]

The distance between the slits and the screen:

[tex]L = 302\:cm = 3.02\:m[/tex]

Distance between 2 slits: [tex]d = 0.00025\:m[/tex]

[tex](n-\frac{1}{2})\lambda=\frac{d(x_n)}{L}[/tex]

For fifth minima, n = 5... so we have:

[tex]x_{5n}=\frac{9\lambda L}{2d}[/tex]

For central maxima, n = 0... so we have:

[tex]x_{0}=\frac{n\lambda L}{d}=0[/tex]

So the distance from central maxima to 5th minimum is:

[tex]\frac{9\lambda \:L}{2d}-0=0.033[/tex]   (Putting the values, we get):

[tex]\Rightarrow \lambda = 6.1\times 10^{-7}\:m[/tex]

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Answer:

Time, t = 3.2 ms

Explanation:

It is given that,

Mass of basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s

Final velocity, v = 3.85 m/s

Average force acting on the ball, F = 72.9 N

We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,

[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms[/tex]

So, the ball is in contact with the floor for 3.2 ms.

Explanation:

Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),

the conditions can be summarized bellow

point A

initial velocity= 0

final velocity=0

P.E= Max

K.E= 0

point B

initial velocity= maximum

final velocity=maximum

P.E=K.E

point C

initial velocity= min

final velocity=min

P.E= 0

K.E= max