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3 years ago

5

A .2kg Basketball is pitched with a velocity or 40 m/s and then bed and into the picture with a velocity of 60 m/s. What is the

change in the balls momentum?

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

40kgm

Explanation:

∆p = m(v - u)

= 2(60 - 40)

= 2 × 20

= 40kgm/s

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Which situation is work not being done? A) A bookcase is slid across carpeting. B) A stack of books is carried at waist level ac

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Answer:

B) A stack of books is carried at waist level across a room

Explanation:

Work is defined as:

$W=Fd cos \theta$

where

F is the force applied

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

From the formula, we see that the work done is zero when the force and the displacement are perpendicular to each other. Let's now analyze each situation:

A) A bookcase is slid across carpeting. --> work is done, because the force that pushes the bookcase is in the same direction of the displacement

B) A stack of books is carried at waist level across a room. --> no work is done, because the force to carry the book is vertical, while the displacement of the books is horizontal

C) A chair is lifted vertically with respect to the floor. --> work is done, because the force that lifts the chair is vertical, and the displacement is vertical as well

D) A table is dropped onto the ground. --> work is done, because the force of gravity (that makes the table falling down) is vertical and the displacement of the table is also vertical.

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3 years ago

What will happen if the two plungers are pressed together firmly? Explain your answer.

steposvetlana [31]

Answer:

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3 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

b)

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

c)

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

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3 years ago