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3 years ago

By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)

Physics

1 answer:

Fofino [41]3 years ago

6 0

Gravity on the surface of sun is given as

$g = \frac{GM}{R^2}$

here we know that

$M = 1.98 \times 10^{30} kg$

$R = 6.95 \times 10^8 m$

now we will have

$g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}$

$g = 273.4 m/s^2$

now we need to find the ratio of weight on surface of sun and on surface of Earth

$R = \frac{mg_{sun}}{mg_{earth}}$

$R = \frac{273.4}{9.8} = 28 times$

so weight will increase by 28 times

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ASAP

scoundrel [369]

Answer:

Explanation:

Hooke's law! F(spring)=-kx

There's no tricky square law here. The spring constant doesn't change, only x (distance stretched) changes. Therefore, if distance is halved, Force will be halved.

5 0

3 years ago

An RLRL circuit with with a 1212-ΩΩ resistor and a 0.06−H0.06−H inductor carries a current of 1A1A at t=0t=0, at which time a vo

tankabanditka [31]

Answer:

The equation of current is $I=0.86 cos \left (120 t + \frac{\pi}{2} \right )$

Explanation:

Resistance, R = 12 ohm

Inductance, L = 0.06 H

E (t) = 12 cos (120 t)

Compare with the standard equation,

$E=E_{0}cos (2\pi ft)$

$2\pi ft = 120 t \\\\\\w = 2\pi f = 120 rad/s$

So, the inductive reactance is

XL = w L = 120 x 0.06 = 7.2 ohm

The impedance of the circuit is

$Z =\sqrt{12^2+7.2^2}\\\\Z = 14 ohm$

The current leads by 90degree so the equation of current is

$I=\frac{Eo}{Z} cos \left (120 t + \frac{\pi}{2} \right )\\\\I=\frac{12}{14} cos \left (120 t + \frac{\pi}{2} \right )\\\\I=0.86 cos \left (120 t + \frac{\pi}{2} \right )$

4 0

3 years ago

A woman falls to the ground while wearing a parachute. The air resistance of the parachute is 500N. If the woman falls at a cons

alina1380 [7]

the answer is less than 500 n

6 0

3 years ago

If a rock weighing 2,200 N is dropped from a height of 15 m, what is its kenetic energy just before it hits the ground

ycow [4]

Kinetic energy of the rock just before it hits the ground=KE=33000 J

Explanation:

Weight= 2200N

mg=2200

m(9.8)=2200

m=224.5 kg

initial velocity=0

final velocity =V

using kinematic equation V²=Vi²+2gh

V²=0+2 (9.8)(15)

V=17.1 m/s

now kinetic energy= 1/2 mV²

KE= 1/2 (224.5)(17.1)²

KE=33000 J

Thus the kinetic energy of the rock just before it hits the ground=33000 J

7 0

3 years ago

2. How far will a bird have traveled in 13 seconds if it maintained a speed of 31 m/s?

Flura [38]

speed of bird = 31 m/s

time = 13 s

since here bird maintains its speed to be constant

so here the distance traveled is given by

$d = v*t$

$d = 31 * 13$

$d = 403 m$

so bird will travel 403 m distance while maintaining its speed to be constant

average speed is the ratio of total distance covered and total time

here Ian walk 2 km then 0.5 km and then 2.5 km so the total distance is given as

$d = d_1 + d_2 + d_3$

$d = 2 + 0.5 + 2.5$

$d = 5 km$

total time of his full motion is

$t = 3 hours$

so the average speed of the motion is

$v = \frac{d}{t}$

$v = \frac{5}{3} = 1.67 km/h$

so here average speed of Ian is 1.67 km/h

Initial speed of the truck is 6 m/s

after 4 s interval of time

final speed of the truck is 14 m/s

so the acceleration is rate of change in velocity

so it is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{14 - 6}{4}$

$a = 2 m/s^2$

so here truck accelerated with 2 m/s^2

Here car starts from rest so its initial speed is 0

after 2.5 s interval of time

final speed of the car is 15 m/s

so the acceleration is rate of change in velocity

so it is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{15 - 0}{2.5}$

$a = 6 m/s^2$

so here car accelerated with 6 m/s^2

a bag full of trash is dropped from the balcony

Here we can say when bag is in air

the net force on the bag is only due to its weight

so here we can use Newton's II law to find the acceleration of bag

$F = ma$

$mg = ma$

$a = g = 9.8 m/s^2$

so here bag will have constant acceleration during its whole motion as we can see that there is no change in the force and hence its acceleration always remains constant

3 0

4 years ago