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shusha [124]
4 years ago
7

An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

he center of the circle, how much work does this force do as the object moves? 1. W = F (2 π r) for a circle radius of r. 2. W = F (π r2 ) for a circle radius of r. 3. Zero 4. W = m v2 r for a circle radius of r. 5. None of these
Physics
1 answer:
Vlad1618 [11]4 years ago
4 0

Answer:5

Explanation:

Given

speed of object v=1\ m/s

radius of circle r=1\ m

Force towards the center F=1\ N

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

W=F\cdot s\cos 90

W=0

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Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (
alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b)E_1=\dfrac{\sigma}{\epsilon_0}

c)E_2=\dfrac{\sigma}{\epsilon_0}

Explanation:

Let \sigma be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field  so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0

b)The Electric Field to the right of the sheets

E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

c)The Electric Field to the left of the sheets

E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}

3 0
3 years ago
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m^2 . The wave is incident
Alchen [17]

Answer:

47 mW

Explanation:

The average value of the Poynting vector, S = 0.939 W/m² = Intensity of wave, I

S = I S

Also, I = P/A where P = Et, P = power of electromagnetic wave, E = energy of electromagnetic wave in time t and t = time = 1 min = 60 s and A = area = lb since the electromagnetic waves falls on area equal to that of a rectangle.

So, S = Et/A

E = SA/t

= Slb/t

= 0.939 W/m² × 1.5 m × 2.0 m/60 s

= 2.817 W/60 s

= 0.047 W

= 47 mW

So, 47 mW of electromagnetic energy falls on the area in 1.0 minute.

4 0
3 years ago
What is the kinetic energy of a penguin with a mass of 8 kg that is running at a speed<br> of 3 m/s?
kumpel [21]

mass (m) = 8kg
velocity (v) = 3 m/s²

Ke = ½mv²

ke = ½ mv²
½ × 8 × 3²
½ × 24²
12
8 0
3 years ago
What is the dimension of area​
never [62]

Thus, area is the product of two lengths and so has dimension L2, or length squared.

3 0
3 years ago
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0
3 years ago
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