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masha68 [24]
3 years ago
14

A 4.89 μC test charge is placed 4.10 cm away from a large, flat, uniformly charged nonconducting surface. The force on the charg

e is 321 N. The charge is then moved 2.00 cm farther away from the surface. What is the force on the test charge now?
Physics
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer:

The new force on the test charge is 145.02N

Explanation:

Force on a unit positive charge can be calculated using coulomb's law.

F =Kq²/r²

Where

F is the force on the charge = 321 N

K is a constant = 8.99 X10⁹ Nm²/C²

q is a test charge = 4.89 μC  = 4.89 X10⁻⁶ C

r is the distance between the charge and the surface = 4.1cm

F ∝1/r²

Force on the charge is inversely proportional the square of distance between the charge and the surface.

Fr² = constant

F₁r₁² = F₂r₂²

F₂ = F₁r₁²/r₂²

If the charge is then moved 2.00 cm farther away from the surface;

The new distance becomes 4.10 cm + 2.00 cm = 6.1 cm

Therefore, r₂ = 6.1 cm, r₁ = 4.1 cm, F₁ =321 N

F₂ = (321 X4.1²)/(6.1²)

F₂ = 145.02N

Therefore, The new force on the test charge is 145.02N

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Sholpan [36]

Answer:

Explanation:

When a person jumps into a swimming pool and is uninjured while the same person is injured when he jumped from the same height onto the concrete floor, this can be explained by the impulse-momentum concept.

When a person jumps into the swimming pool, water provides cushioning and gradually decreases the velocity of a person with a considerable amount of time. While on the other hand concrete floor does not provide the cushioning effect as it is rigid.

So, force is imparted at a short amount of time causing more injury compared to the swimming pool case.

6 0
4 years ago
If one knows only the constant resultant force acting on an object and the time during which this force acts
Alenkinab [10]

Explanation:

  • We know that Newton's second law of motion gives the measure of external force acting on an object. It says that the rate of change of linear momentum of an object is directly proportional to the external applied force. It is given by :

F=\dfrac{\Delta p}{\Delta t}

Since, p = mv m is mass and v is velocity.

F=\dfrac{m(v-u)}{ t}\\\\F\times t=m(v-u)

So, if one knows only the constant resultant force acting on an object and the time during which this force acts, one can determine the change in momentum of the impulse.

5 0
3 years ago
Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot
frozen [14]

Answer:

0.285

Explanation:

Given two forces of different magnitude, it is important to note that the product of normal force and coefficient of kinetic friction should be equal to the sum of these two forces at equilibrium. Therefore, this can be Mathematically expressed as:

N/\mu_k=F_1+F_2\\\\

where N is normal force,\mu is coefficient of static friction, F is force and subscripts 1 and 2 represent larger and smaller magnitude forces respectively.  Making \mu the subject of the formula then

\mu_k=\frac{F_1+F_2}{N}

Since normal force N is also given by mg where m is mass of object and g is acceleration due to gravity then substituting N with mg we obtain that

\mu_k=\frac{F_1+F_2}{mg}  and substituting the figures given in the question, taking g as 9.81 we obtain that

\mu_k=\frac { 430 N+380 N}{290\times 9.81}=0.285

Hence,the coefficient of kinetic energy is 0.285 as calculated

4 0
4 years ago
A 2.0-m wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of t
konstantin123 [22]

Answer:

The force experienced  is 0.6 N

Explanation:

Given data

length of wire L= 2 m

current in wire I= 0.6 A

magnetic field B= 0.5

The force experienced can be represented as

F= BIL

F= 0.5*0.6*2\\\F= 0.6 N

7 0
3 years ago
In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal
storchak [24]

Answer:

c. Both forces have equal magnitudes.

Explanation:

According to Newton´s 3rd Law, the force exerted by a body on another one , is equal and opposite to the one that the second object exerts on the first one.

This does not depend on the masses of the objects, so it doesn´t matter that m₁ be greater than m₂.

As examples of this, we can mention the gravity force, the electrostatic force, the normal force, etc.

So, in this case, taking into account only the magnitudes, we can say:

F₁₂ = F₂₁

7 0
3 years ago
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