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3 years ago

14

Jamal plugged his radio into the wall. The radio's plug had copper wires surrounded by rubber. The rubber protects Jamal from___

___?

2 answers:

Mariana [72]3 years ago

3 0

The rubber protects him from being electrocuted by the flow of current going through the plug.
Hope this helped!!

aliya0001 [1]3 years ago

3 0

The rubber surrounding the wires protects Jamal from electrocution.

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Help with this please.

gregori [183]

Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>

8 0

3 years ago

A system gains 757 kJ757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ.+176 kJ. How much w

Alecsey [184]

Answer:

581 kJ, work was done by the system

Explanation:

According to the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released)

W is the work done by the system (positive if done by the system, negative if done by the surrounding)

In this problem,

$Q=+757 kJ$

$\Delta U = +176 kJ$

Therefore the work done by the system is

$W=Q-\Delta U=757 kJ-176 kJ=+581 kJ$

And the positive sign means the work is done BY the system.

5 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

3 years ago

an object with a mass of 2.4 kg has a force of 12.6 N applied to it. What is the resulting acceleration of the object? WITH PROO

Scorpion4ik [409]

Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2

3 0

3 years ago

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

$F_2 = \frac{K2q_12q_2}{r^2} = \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1$

Therefore, the force between them quadruples

4 0

3 years ago