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3 years ago

Convert 55 miles into inches

Physics

2 answers:

anastassius [24]3 years ago

8 0

Answer:

What I got was a big number but it got around 3.485e+6

Explanation:

frosja888 [35]3 years ago

6 0

Answer:

3.485e+6 inches. hope this helps

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

aleksklad [387]

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force T is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T = mg - mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60 { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

5 0

3 years ago

a mass of 10kg is placed on a horizontal table with coefficient of friction is 0.5. if the mass is static, determine weight of t

zloy xaker [14]

Answer:

Explanation:

5 0

3 years ago

7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained

marta [7]

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

mass of the metal hammer, m = 1.0 kg
speed of the hammer, v = 5.0 m/s
specific heat capacity of iron, 450 J/kg⁰C

The increase in temperature of the metal hammer is calculated as follows;

$Q = K.E\\\\mc \Delta T = \frac{1}{2} mv^2\\\\\Delta T = \frac{v^2}{2 c}$

where;

c is the specific heat capacity of the metal hammer

Assuming the metal hammer is iron, c = 450 J/kg⁰C

$\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C$

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

Learn more about heat capacity here: brainly.com/question/16559442

8 0

3 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

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3 years ago

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Kipish [7]

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3 years ago

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