Answer:
Liquids
Explanation:
Liquids take up the shape of the container it is poured into but will never change its volume.
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:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.
A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.15 s
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.
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20 C and the convection coefficient at the outer tube surface is 11 W/m2 K. Estimate the heat loss per unit length of tube.
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.
Calculation of the heat loss: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
Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at a constant velocity, how hard and in which direction is team B pulling?
A. 2500 N to the left
B. 5000 N to the right
C. 2500 N to the right
D. 5000 N to the left
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.
Choose the friction which opposes the relative motion between surfaces in motion a.Static friction b.Kinetic friction c.Sliding friction d.Both kinetic and sliding friction
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.
when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how much time was it in contact with the floor?
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.