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4 years ago

Explain objects in space motion. How do they move? What affects their motion?

1 answer:

5 0

Objects in space follow the laws or rules of physics, just like objects on Earth do. Things in space have inertia. That is, they travel in a straight line unless there is a force that makes them stop or change. The movement of things in space is influenced by gravity.

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Which force represented by the arrow at B? a. force of friction b. force of gravity c. tension force d. normal force

r-ruslan [8.4K]

Answer:

A. Force of friction

Explanation:

Find the diagram attached

From the diagram we can see that the force B is acting up on the body along the plane. This force is the force that prevents the body from sliding down the plane. Originally, the body on the line will tend to slide down the incline even without application of a force hence the reason of an opposing force that must act in the opposite direction is counter this motion to make the body be in a state of equilibrium. Hence the required force is force of friction since it is am opposing force acting in the direction opposite to the moving force acting on the body.

5 0

3 years ago

A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th

Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0

3 years ago

A 27.0-g object moving to the right at 21.5 cm/s overtakes and collides elastically with a 11.0-g object moving in the same dire

7nadin3 [17]

Answer:

final speed of the first object: 17.73 cm/s

final speed of the second object: 24.23 cm/s

Explanation:

Data:

mass of the first object, m1 = 27.0 g

mass of the second object, m2 = 11.0 g

initial speed of the first object, v1i = 21.5 cm/s

initial speed of the second object, v2i = 15 cm/s

final speed of the first object, v1f = ? cm/s

final speed of the second object, v2f = ? cm/s

Elastic collisions formula:

v1f = (m1 - m2)/(m1 + m2) * v1i + 2*m2/(m1 + m2) * v2i

v1f = (27 - 11)/(27 + 11) * 21.5 + 2*11/(27 + 11) * 15

v1f = 17.73 cm/s

v2f = 2*m1/(m1 + m2) * v1i + (m2 - m1)/(m1 + m2)*v2i

v2f = 2*27/(27 + 11) * 21.5 + (11 - 27)/(27 + 11)*15

v2f = 24.23 cm/s

8 0

3 years ago

An extended period of well-below-average temperatures is known as a

jasenka [17]

Well, I think it kinda depends on how long the "extended period" is.

If the extended period is a week, a month, or a few months,
then I'd call it a 'cold wave'.

If the extended period is like 100 years or more, then you're
starting to talk 'ice age'.

3 0

3 years ago

Read 2 more answers

Each milligram of glucose has the same amount of energy available to do work. The series B test tubes produced more bacteria per

yKpoI14uk [10]

Answer:

The series A test tube has some left amount of glucose left in it.

Explanation:

Let's assume that a fixed amount of glucose is synthesized, for the fixed quantity the bacteria produced in A and B be x and y respectively,

Therefore, the condition on x and y is, y > x as the no. of bacteria present in B is greater.

As a result B would require a greater amount of energy for its functioning, these energy would be derived from the already fixed amount of glucose present.

A test tube would also require the energy for its x number of bacteria, but it is less than that of B.

Therefore, there would be some unused glucose left in Test Tube Series A which has unused energy.

4 0

3 years ago