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4 years ago

Based on this electric field diagram, which statement best compares the charge of A with B?

1 answer:

4 0

Based on this electric field diagram, the statement which best compares the charge of A with B is "A is negatively charged and B is positively charged. The charge on A is greater than that on B".

<u>Answer:</u> Option A

<u>Explanation:</u>

The charge is quantized represented as elementary charge, about 1.602×10−19 coulombs. Their are two kinds of electric charging: positive and negative (usually transported, separately, by protons and electrons). Like charges repel each other, while attraction occurs among unlike charges. An entity without net charge is considered neutral. If a piece of matter comprises more electrons than protons, it has a negative charge, when there are fewer, it'll have a positive charge and when there are equal amounts, this will be neutral.

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Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

4 years ago

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

3 years ago

The acceleration of the body in the graph given below is____

Art [367]

Acceleration = -12 / 8 = - .... m/s^2

5 0

3 years ago

find the coefficient of kinetic friction for a 10 kg box being dragged steadily across the surface with a force of 2.0 Newtons

Mazyrski [523]

Answer:

0.02

Explanation:

Force, $F=\mu_k N$ where N is normal reaction and coefficient of kinetic friction is $\mu_k$ . Also, N is equivalent to mg ie N=mg where m is mass of an object and g is acceleration due to gravity. Making $\mu$ the subject of the formula then