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3 years ago

An astronaut goes to Mars to do some experiments. Explain why her mass stays the same but her weight changes.

1 answer:

6 0

Because mass does not change from place to place but weight does change from place to place... why? because weight is the amount of gravitational force on an object and mass is the amount of matter in an object. mars has less gravitational force so an object will weigh less than it really weighs there

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Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t

MissTica

Answer: Velocity, Gravity, and Momentum

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3 years ago

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The water in the plumbing in a house is at a gauge pressure of 300,000 pa. What force does this cause on the top of the tank ins

liraira [26]

Answer:

60 000 N

Explanation:

1 pa = 1 N/m^2

you have 300 000 of these = 300 000 N /m^2

but only an area of .2 m^2

300 000 N / m^2 * .2 m^2 = 60 000 N

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2 years ago

A step index fiber has a numerical aperture of NA = 0.1. The refractive index of its cladding is 1.465. What is the largest core

Vladimir [108]

Answer:

diameter = 9.951 × $10^{-6}$ m

Explanation:

given data

NA = 0.1

refractive index = 1.465

wavelength = 1.3 μm

to find out

What is the largest core diameter for which the fiber remains single-mode

solution

we know that for single mode v number is

V ≤ 2.405

and v = $\frac{2*\pi *r}{ wavelength} NA$

here r is radius

so we can say

$\frac{2*\pi *r}{ wavelength} NA$ = 2.405

put here value

$\frac{2*\pi *r}{1.3*10^{-6}} 0.1$ = 2.405

solve it we get r

r = 4.975979 × $x^{-6}$ m

so diameter is = 2 × 4.975979 × $10^{-6}$ m

diameter = 9.951 × $10^{-6}$ m

3 0

3 years ago

What forces affect a toy car as it moves down a ramp?

Amanda [17]

Answer:

gravity

Explanation:

3 0

3 years ago

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A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc

____ [38]

Answer:

Part a)

$a = 0.36 m/s^2$

Part b)

$a = -1.29 m/s^2$

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

$F_x = Fcos25$

$F_x = 80 cos25 = 72.5 N$

$F_y = Fsin25$

$F_y = 80 sin25 = 33.8 N$

now in vertical direction we can use force balance for the box to find the normal force on it

$F_n + F_y = mg$

$F_n = (25)(9.81) - 33.8$

$F_n = 211.45 N$

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

$F_k = (0.300)(211.45) = 63.44 N$

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

$f_k = (0.300)(207.7) = 62.3 N$

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

7 0

4 years ago