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3 years ago

7

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . What is the electric force acting on the

charge?

2 answers:

Darina [25.2K]3 years ago

3 0

The force of a charged particle in an electric field is given by F=qE, where q is the charge and E is the field in N/C. This gives 0.9N

otez555 [7]3 years ago

3 0

Answer:

0.9

Explanation:

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An object is placed 9.5 cm in front of a convex spherical mirror. Its image forms 3.2 cm behind the mirror. What is the radius o

o-na [289]

Using the mirror formula.

1/v + 1/u = 1/f

1/9.5 + 1/3.2 = 1/f

1/f = 3.2 + 9.5 / 9.5 * 3.2

1/f = 4.82 cm

Radius = 2f

Radius = 2 x 4.82

Radius = 9.64 cm

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4 years ago

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What happens to the intensity of solar energy as latitude increases

lora16 [44]

As the latitude increases the intensity of solar energy decreases

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4 years ago

A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'

Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = $\frac{1}{2}$ I ω^2 + $\frac{1}{2} mv^{2}$

where

I = $\frac{1}{2} mr^{2}$ and ω = $\frac{v}{r}$

thus,

KE = $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ ( $\frac{v}{r}$ )^2 + $\frac{1}{2} mv^{2}$

= $\frac{1}{2}$ $\frac{1}{2} mr^{2}$ $\frac{v^{2} }{r^{2}}$ + $\frac{1}{2} mv^{2}$

= $\frac{1}{4} mv^{2}$ + $\frac{1}{2} mv^{2}$

= $\frac{3}{4} mv^{2}$

= $\frac{3}{4} (0.46) (1.1)^{2}$

= 0.42 J

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3 years ago

A 115-turn circular coil of radius 2.71 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi

tester [92]

Answer:

80.6 mV

Explanation:

Parameters given:

Number of turns, N = 115

Radius of coil, r = 2.71 cm = 0.0271m

Time taken, t = 0.133s

Initial magnetic field, Bin = 50.1 mT = 0.0501 T

Final magnetic field, Bfin = 90.5 mT = 0.0905 T

Induces EMF is given as:

EMF = [(Bfin - Bin) * N * A] / t

EMF = [(0.0905 - 0.0501) * 115 * pi * 0.0271²] / 0.133

EMF = (0.0404 * 115 * 3.142 * 0.0007344) / 0.133

EMF = 0.0806 V = 80.6 mV

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3 years ago

A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16

erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle $\phi$ = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

$V =\pi D( \frac{N}{60} )$

$V = 3.14 (0.034) \frac{(1650)}{60}$

$V = 2.93 \frac{m}{s}$

Form factor for the pinion gear is

Y = 0.303

Now

$K_{v} = \frac{6.1 +0.303}{6.1} = 1.049$

Force on gear tooth

$F = \frac{P}{V}$

$F = \frac{1200}{2.93}$

F = 408.73 N

Now the bending stress is given by the formula

$\sigma = \frac{K_{v} F}{m b y}$

$\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}$

$\sigma$ = 35.38 M pa

This is the value of bending stress on the pinion

8 0

3 years ago