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sertanlavr [38]
3 years ago
6

A negative charge of -0.550 m exerts an upward

68681816">[email protected] force on an unknown charge that is located 0.300 m directly below the first charge. What are (a) the value of the unknown charge (magnitude and sign); (b) the magnitude and direction of the force that the unknown charge exerts on the [email protected] charge
Physics
1 answer:
almond37 [142]3 years ago
8 0

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

q₂=10.9μC.

Hence the second charge is +10.9μC

b. From the rule of charges which state that like charges repel and unlike charges attract, we can conclude that the two above charges will attract since they are unlike charges. Hence the direction of the force will be downward into the second charge and the magnitude of the force will remain the same as 0.600N

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rodikova [14]

Answer:

See the answers below.

Explanation:

The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.

Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]

The fuse can be calculated by knowing the amperage.

P=V*I

where:

P = power = 350 [W]

V = voltage = 240 [V]

I = amperage [amp]

Now clearing I from the equation above:

I=P/V\\I=350/240\\I=1.458[amp]

The fuse should be larger than the current of the circuit, i.e. about 2 [amp]

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3 years ago
Lemonade is best described as _______. A. an element B. a compound C. a mixture D. a pure substance
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C is the answer because it mix and water combining together
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if a car travels 30 miles north for one half hour, 50 miles east for one hour, and 30 miles south for 30 minutes, what is the to
egoroff_w [7]
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You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)
Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

a = g = 9.8 m/s^2

Part b)

Now from kinematics equation we will have

y = v_i t + \frac{1}{2} at^2

now we have

y = 25 m

so from above equation

25 = 0 + \frac{1}{2}(9.8 )t^2

t = 2.26 s

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

v_x = vcos\theta

v_x  = 20 cos65 = 8.45 m/s

v_y = vsin\theta

v_y = 20 sin65 = 18.13 m/s

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

v_f - v_y = at

0 - 18.13 = (-9.8) t

t = 1.85 s

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5 0
3 years ago
A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho
creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano (m) = 190 kg

Inclined angle (\theta) = 18 degree

Considering gravity, g = 9.8 ms^-^2

And

Using, sin(18) =0.30 and cos(18)=0.95

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as  \sum F=0 .

⇒ F+mgsin(\theta) =0

⇒ F=-mgsin(\theta)

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass (m) and angle (\theta) .

⇒ F=190\times 9.8\times sin(18)

⇒ F=575.38 N

b.

When the force is parallel to the floor.

⇒ Fcos(\theta)=mgsin(\theta)

⇒ F=\frac{mgsin(\theta)}{cos(\theta)}

⇒ Plugging the values.

⇒ F=\frac{190\times 9.8\times sin(18)}{cos(18)}

⇒ F=605 N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

6 0
3 years ago
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