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3 years ago

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Light travels at a speed of 2.998*108 m/s. Light takes approximately 3.25 minutes to travel from the Sun to reach a planet. Calc

ulate the distance from the Sun to this planet in meters. Give your answer to 0 decimal places.

1 answer:

ANTONII [103]3 years ago

5 0

Answer:

585×10⁸ m

Explanation:

Distance = rate × time

d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)

d = 585×10⁸ m

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A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the

Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating system = 6.74 J

We know that energy is given by

(a) $Ke=\frac{1}{2}mv_{max}^2$

$6.74=\frac{1}{2}\times 0.267\times v_{max}^2$

$v_{max}=7.1m/sec$

(b) Now $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec$

We know that $\omega =\sqrt{\frac{k}{m}}$

$35.68=\sqrt{\frac{k}{0.267}}$

$k=339.9N/m$

(c) We know that energy is given by

$E=\frac{1}{2}KA^2$

$6.74=\frac{1}{2}\times 339.9\times A^2$

$A=19.91cm$

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3 years ago

Calculate the pressure of water in a well if the depth of the water is 10m

Kryger [21]

Answer:

P= phg

so:- the height is 10m

density of water 1000 kg/m3

gravity is 9.8m/s2

P=1000 kg/m3*10m*9.8m/s2

=98000Pa

=98KPa

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2 years ago

Is this true or false the kinetic energy of a freely falling object decreases as it begins to fall

Naddika [18.5K]

The kinetic energy of an object is the energy of the object as it moves. It has more kinetic energy the faster it moves. Because a falling object is increasing in speed, it would also increase in kinetic energy. Hope this helps! :)

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3 years ago

Read 2 more answers

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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3 years ago

A circuit has a resistance of 10ohmsband a current of 42 amps.caculate the voltage

Anon25 [30]

R=U/I so
U=RxI
U= 10 x 42
U= 420 volts

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3 years ago