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3 years ago

4-08 t 1 - / S P = 1 / P = S 1 S= P Def. of Speed

Physics

1 answer:

nordsb [41]3 years ago

8 0

Answer:

what do I answer? not enough information

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A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it

joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

$U = F.s$

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

$30 = F(1.5)$

$F = 20 N$

6 0

3 years ago

The perimeter of a rectangle can be found using the equation P = 2L + 2W, where P is the perimeter, L is the length, and W is th

sweet [91]

Answer:

The perimeter of the rectangle is 60 units.

Explanation:

The perimeter of a rectangle is given by the equation P = 2L + 2W. With W = 12 units and L = 18 units, substituting the values in the equation above:

P = 2(18 units) + 2(12 units)

P = 36 units + 24 units

P = 60 units

8 0

3 years ago

What affects the vertical velocity of a projectile over time?<br>---

kobusy [5.1K]

The vertical velocity is affected by the acceleration of gravity (ignoring the effects of air resistance usually)

6 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,

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

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

$\%error = 2.7%$

8 0

3 years ago

The lever allows Jeff 150 pounds to lift a much greater weight 600 pounds because A) the larger force is applied over a longer d

kvasek [131]

Jeff uses all of his weight to lift the 600 lbs. C is your answer i do believe

4 0

3 years ago

4-08 t 1 - / S P = 1 / P = S 1 S= P Def. of Speed​

4-08 t 1 - / S P = 1 / P = S 1 S= P Def. of Speed