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3 years ago

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What is the weight of a 435 kg object on earth? 957.98 N 957.98 N None of these answers are correct. None of these answers are c

orrect. 197.53 N 197.53 N 7.20 N 7.20 N 7203 N

1 answer:

mixer [17]3 years ago

7 0

Answer:

none of these option are correct....

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The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling

Crank

270/70^2 = x/80^2

Cross multiply

270 (6400) = 4900x

x = 270(6400)/4900

352 and 32/49 feet

Hope this helps

4 0

3 years ago

The human nerve cells have a net negative charge and the material in the interior of the cell is a good conductor. if the cell h

mash [69]

The magnitude and direction (inward or outward) of the net flux through the cell boundary is - 0.887 wb.m².

<h3>What is flux?</h3>

Flux describes any effect that appears to pass or travel through a surface or substance.

The magnitude and direction (inward or outward) of the net flux through the cell boundary is calculated as follows;

Ф = Q/ε

where;

Q is net charge
ε is permittivity of free space

Φ = (-7.85 x 10⁻¹²)/(8.85 x 10⁻¹²)

Φ = - 0.887 wb.m²

Learn more about flux here: brainly.com/question/10736183

#SPJ1

6 0

2 years ago

Which winds are affected by specific landforms on earth's surface?

Blababa [14]

Hello!

The winds affected by specific landforms on earth's surface are: Local winds.

I hope my answer helped you out! :)

4 0

3 years ago

Torque can cause the angular momentum vector to rotate in UCM. This motion is called ___________.

emmainna [20.7K]

Torque can cause the angular momentum vector to rotate in UCM. This motion is called _Conservation of Angular momentum__________.

Answer:

Conservation of Angular momentum

Explanation:

The motion of an object in a circular path at constant speed is known as uniform circular motion (UCM). An object in UCM is constantly changing direction, and since velocity is a vector and has direction, you could say that an object undergoing UCM has a constantly changing velocity, even if its speed remains constant.

The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.

Key Points

When an object is spinning in a closed system and no external torques are applied to it, it will have no change in angular momentum.

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation.

If the net torque is zero, then angular momentum is constant or conserved.

Angular Momentum

The conserved quantity we are investigating is called angular momentum. The symbol for angular momentum is the letter L. Just as linear momentum is conserved when there is no net external forces, angular momentum is constant or conserved when the net torque is zero. We can see this by considering Newton’s 2nd law for rotational motion:

τ→=dL→dt, where

τ is the torque. For the situation in which the net torque is zero,

dL→dt=0.

If the change in angular momentum ΔL is zero, then the angular momentum is constant; therefore,

⇒

L =constant

L=constant (when net τ=0).

This is an expression for the law of conservation of angular momentum.

Example and Implications

An example of conservation of angular momentum is seen in an ice skater executing a spin, The net torque on her is very close to zero,

because (1) there is relatively little friction between her skates and the ice, and (2) the friction is exerted very close to the pivot point.

Conservation of angular momentum is one of the key conservation laws in physics, along with the conservation laws for energy and (linear) momentum. These laws are applicable even in microscopic domains where quantum mechanics governs; they exist due to inherent symmetries present in nature.

7 0

3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago