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3 years ago

What is the distance magnification formula

Physics

1 answer:

damaskus [11]3 years ago

7 0

Does that help I hope IT does you probably just have to write the 1st sentence.

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The graph of an object's position over time is a horizontal line and y is not equal to 0. What must be true abou

frozen [14]

Answer:D: the velocity is zero

Explanation:

7 0

2 years ago

g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k

Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

$(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s$

Now, we will apply the law of conservation of momentum:

$m_1v_1 = m_2v_2$

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

$(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s$

Now, we again use the third equation of motion for the upward motion of the ball:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

$(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\$

6 0

2 years ago

In the following diagram, the voltage is 1.5 volts and the resistance is 0.35 ohms. Use Ohm's Law to determine the current in th

Angelina_Jolie [31]

Answer:

I = 4.28 [amp]

Explanation:

To solve this type of problems we must have knowledge of the law of ohm, which tells us that the voltage is equal to the product of resistance by current.

Initial data:

v = 1.5 [volt]

R = 0.35 [ohms]

v = I * R

therefore:

I = 1.5 / 0.35

I = 4.28 [amp]

5 0

2 years ago

A 1 pound bucket rests on a table. What is the support force exerted on the bucket by the table? What is the support force when

katrin2010 [14]

This question involves the concepts of equilibrium and Newton's third law of motion.

The support force will be "1 pound" for the empty bucket and the support force will be "6 pounds" after pouring water into it.

According to the condition of equilibrium, the sum of forces acting on a stationary object must be zero. Hence, the support force of the table will be equal to the total mass of the bucket.
According to Newton's Third Law of Motion every action force has an equal but opposite reaction force. Hence, the support force will be a reaction force to the weight of the bucket.

Therefore, the support force in each case will be equal to the total mass of the bucket:

Case 1 (empty bucket):

<u>support force = 1 pound</u>

<u></u>

Case 1 (water poured):

support force = 1 pound + 5 pound

<u>support force = 6 pound</u>

<u></u>

Learn more about equilibrium here:

brainly.com/question/9076091

8 0

2 years ago

Please help me

melamori03 [73]

<h2>Question:</h2>

An automobile is driving uphill. Which form of energy is not involved in this process?

<h2>Choosing:</h2>

electromagnetic

potential

kinetic

chemical

<h2>Answer:</h2>

<u>Electromagnetic</u><u> </u>

<h3><u>#READINGHELPSWITHLEARNING</u><u> </u></h3><h3><u>#CARRYONLEARNING</u><u> </u></h3><h3><u>#STUDYWELL</u><u> </u></h3>

7 0

3 years ago