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Mashcka [7]
3 years ago
7

10 rectangular object's mass from greatest to least. You can choose books, sandwiches, phones, pictures - as long as the shape i

s a rectangle. Then rank the objects from the one with the heaviest mass to the one with the lightest mass.
Physics
1 answer:
Greeley [361]3 years ago
7 0

Answer: 1 is phone 2 is sandwich, Last is picture.

Explanation: I hoped That Helped !!

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Jack has two boxes. One is 148g and one is 78g. If jack pushes both boxes with the same amount of force, which will accelerate f
Step2247 [10]
The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.
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3 years ago
What is someone who leaves there country to settle in another?
Eddi Din [679]

Answer:

an immigrant

Explanation:

8 0
3 years ago
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Each of the different colors that make up white light has a different ___________________.
Verdich [7]

Answer:

Wavelength?

Explanation:

7 0
2 years ago
A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic
Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light c= 2.99793\times10^{8}\ m/s

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}

Put the value into the formula

\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}

\Delta l=1\sqrt{1-(0.96)^2}

\Delta l=0.28\ m

We need to calculate the time the stick take to move

Using formula of time

t=\dfrac{\Delta l}{v}

Put the value into the formula

t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}

t=9.72\times10^{-10}\ sec

t=0.97\ ns

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0
3 years ago
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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