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3 years ago

If you are in an elevator that speed up then the apparent weight is

Physics

1 answer:

Sholpan [36]3 years ago

3 0

Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

$W_{net}$ = mg + ma

where: mg is the weight of the object, and ma is the apparent weight.

Apparent weight (ma) = $W_{net}$ - mg

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hurricane camille was so strong that it broke the instruments used to determine wind speed. how were scientists able to determin

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Possibly by the damage done compared to the previous ones.

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4 years ago

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In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

ANTONII [103]

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3 years ago

In air, the 57 GHz radio wave has a shorter<br> wavelength compared to the 64 GHz radio wave.

dalvyx [7]

Answer:

False

Explanation:

Higher frequency signals have shorter wavelengths. Hence, the 64 GHz signal will have a shorter wavelength than the 57 GHz signal.

Frequency and wavelength are inversely proportional meaning as frequency increases wavelength decreases and as frequency decreases wavelength increases.

Cheers.

6 0

3 years ago

You are riding a bicycle. You apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg. What is the

Ymorist [56]

<u>Answer</u>

1.667 m/s²

Using the Newton's second law of motion;

force is directly proportional to the rate of change of momentum and it takes place in the direction of force.

i.e F = change in momentum

F = (mv - mu)/t

=m(v-u)/t

= ma

∴ F =ma

a = F/m

= 150/90

= 1.667 m/s²

4 0

3 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

4 years ago