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shepuryov [24]
3 years ago
5

If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscil

lations. What can we say, about these springlike bonds when the charge is first, furthest away and second, closest to the source charge
Physics
1 answer:
Hoochie [10]3 years ago
5 0

Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second  and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length  of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>

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A 2.0-N force acts horizontally on a 10-N block that is initially at rest on a horizontal surface. The coefficient of static fri
Nady [450]

Answer:

None of the statement is false - all of them are true

Explanation:

Let's analyze each statement:

- The net force on the block is zero newtons. --> TRUE. In fact, the block is moving across the surface at constant speed: constant speed means zero acceleration, and according to Newton's second law,

F = ma

this also means a net force of zero newtons.

- The frictional force on the block has magnitude 3.0 N. --> TRUE. The horizontal acceleration of the block is zero, so the resultant of the horizontal forces must be zero:

F-F_f = 0

where F = 3.0 N is the horizontal push and F_f is the frictional force. From the equation, we find

F_f = F = 3.0 N

- The coefficient of kinetic friction between the block and the surface is 0.30. --> TRUE. The frictional force is 3.0 N, and its expression is

F_f = \mu_k (mg)

where \mu_k is the coefficient of kinetic friction and (mg)=10 N is the weight of the block. Solving for \mu_k, we find

\mu_k = \frac{F_f}{mg}=\frac{3 N}{10 N}=0.30

- The direction of the total force that the surface exerts on the block is vertically upward. --> TRUE. Since gravity is acting downward, and the block is not accelerating on the vertical direction neither, there must be an equal and opposite force acting upward on the block: and this force is the force exerted by the surface on the block.

- The block is not accelerated. --> TRUE: the block is moving at constant speed, so its acceleration is zero.

7 0
4 years ago
A child’s toy sits on the bottom of a swimming pool in which the water depth is d=1.8 m. To a child standing at the pool edge, t
navik [9.2K]

Answer:

Horizontal distance is 1.28 + 2.04 = 3.32 m

Explanation:

Given data:

from  below figure

Applying Pythagoras theorem

from Snell's law

\frac{sin i}{sin r} = \frac{n_2}{n_1} = \frac{\frac{EF}{AB}}{\frac{DE}{BD}}

1.33 =\frac{ \frac{2.04}{\sqrt{(3.5 -1.8)^2 + 2.04^2}}}{\frac{x}{\sqrt{1.8^2 + x^2}}}

solving for x we have

x^2 = 1.626

x = 1.28

Horizontal distance is 1.28 + 2.04 = 3.32 m

8 0
3 years ago
Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object be?
Serjik [45]

Answer:

I belive it would be ture

Explanation:

It's been a while since I learned this but I think that is right.

6 0
3 years ago
Read 2 more answers
Points A and B lie within a region of space where there is a uniform electric field that has no x- or z-component; only the y-co
liraira [26]

Answer:

(a) Ey is negative

(b) The magnitude of the electric field is E = 171.429 V/m

(c) The potential difference between points B and C is 17.1429 V

Explanation:

(a) Here, we have the potentials given by;

V_A - V_B = +12.0V with point A at y = 8.00 cm and point B at point y = 15.0 cm

where point B is at a higher potential than point A, that is the electric potential is from;

B with y = 15.0 cm to A with y = 8.0 cm which means

E_y decreases as y increases or E_y  is negative.

(b) The magnitude of the electric field is given by

The work done to move a charge from B to A is

W_{BA} = - \Delta U where

\Delta U = U_a -U_b = q_0E(y_b-y_a)

V_{BA} = \frac{\Delta U}{q_0} = \frac{q_0E(y_b-y_a)}{q_0}  = E(y_b-y_a)

∴ E = \frac{V_{BA}}{(y_b-y_a)}

E = \frac{12 \hspace{0.09cm}V}{(0.015\hspace{0.09cm} m -0.008\hspace{0.09cm} m)}

E = 171.429 V/m

(c) Here we have point C x = 5.00 cm and y = 5.00 cm

Therefore we have the distance from B to C given by

y_b-y_c = 15.00 \hspace{0.09cm}cm - 5.00  \hspace{0.09cm}cm = 10.00 \hspace{0.09cm} cm

Where 10.00 cm = 0.01 m

E = V/Δy

Therefore, V = Δy·E

For V_{BC}, Δy = y_b-y_c  = 0.01 \hspace{0.09cm} m and we have,

V_{BC} = E\times (y_b-y_c)

V_{BC} = 171.429\times (0.015-0.005) = 17.1429\hspace{0.09cm}V

7 0
3 years ago
A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy
Nady [450]

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

E_i=\dfrac{mv^2}{2}+mgh

Here , initial velocity is zero .

Therefore , E_i=mgh=40\ J

Now , final energy of the system :

E_f=\dfrac{mv^2}{2}+mg(0)+15\\\\E_f=\dfrac{0.05\times v^2}{2}+15

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

E_i=E_f\\40=\dfrac{0.05v^2}{2}+15\\\\25=\dfrac{0.05v^2}{2}\\\\v=31.62\ m/s

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

3 0
3 years ago
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