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3 years ago

A 750-newton person stands in an elevator that is

1 answer:

8 0

When an elevator is accelerating downward, the normal force is equal to mg-ma (hence you feel a little lighter when accelerating downwards)

Therefore, the upward force of the elevator floor on the person must be less than 750N

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A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block

Helen [10]

Answer: mg/Cosθ

Explanation:

Taking horizontal acceleration of wedge as 'a'

FCosΘ = FsinΘ

F = mass(m) × acceleration(a) = ma

For horizontal resolution g = 0

Therefore,

Horizontal = Vertical

maCosΘ = mgSinΘ

aCosΘ = gSinΘ

a = gSinΘ/CosΘ

Recall from trigonometry :

SinΘ/Cosθ = tanΘ

Therefore,

a = gtanΘ

Normal force acing on the wedge:

mgCosΘ + maSinΘ - - - - (y)

Substitute a = gtanΘ into (y)

mgCosΘ + mgtanΘsinΘ

tanΘ = sinΘ/cosΘ

mgCosΘ + mgsinΘ/cosΘsinΘ

mgCosΘ + mgsin^2Θ/cosΘ

Factorizing

mg(Cosθ + sin^2Θ/cosΘ)

Taking the L. C. M

mg[(Cos^2θ + sin^2Θ) /Cosθ]

Recall: Cos^2θ + sin^2Θ = 1

mg[ 1 /Cosθ]

mg/Cosθ

6 0

3 years ago

PLEASE HELP ME

ra1l [238]

C should be the answer

8 0

3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

At what angle are the electronic and the magnetic wave related in an electromagnetic signal?

VikaD [51]

Answer:

90degrees I'm pretty sure

7 0

3 years ago

Which of the following statements is accurate?

cupoosta [38]

Answer: A and B

Explanation:

The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Because wavelength is the distance between the two successful crest or trough.

Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.

Because amplitude is the measure of maximum displacement from the original position

4 0

3 years ago