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3 years ago

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Three forces acting at a point keep it in equilibrium . If the angle between two of the forces , 3 N and 3 Nis 120°, then the th

ird force
is

1 answer:

Gennadij [26K]3 years ago

8 0

Answer:

see attachments

Explanation:

The force necessary to maintain equilibrium depends on the directions of the two forces that are separated by 120°. If both are pointing toward or away from the point, the third force will be pointing that same direction. It will be 120° from the other two, and the same magnitude (3 N).

If the one of the two forces is pointing toward, and the other is pointing away, then the third force will have magnitude 3√3 N, and will be 30° from one of the other forces. This is illustrated in the 2nd attachment.

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tino4ka555 [31]

Answer:

Mercury:

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Venus:

2.1

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2.2

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1.2

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Explanation:

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2 years ago

What properties of the wave define why it is found within this area of the spectrum?

MatroZZZ [7]

Answer:

Explanation:

Speed and medium are properties of a wave which is in common within an area of the spectrum of visible light. i.e;

Their medium of propagation

They both travel at the same speed ( speed of light )

The properties mentioned above are properties that define that wave is found within an area spectrum for visible light.

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7 0

2 years ago

In this example we will analyze the forces acting on your body as you move in an elevator. Specifically, we will consider the ca

dexar [7]

Answer:

The reading of the scale during the acceleration is 446.94 N

Explanation:

Given;

the reading of the scale when the elevator is at rest = your weight, w = 600 N

downward acceleration the elevator, a = 2.5 m/s²

The reading of the scale can be found by applying Newton's second law of motion;

the reading of the scale = net force acting on your body

R = mg + m(-a)

The negative sign indicates downward acceleration

R = m(g - a)

where;

R is the reading of the scale which is your apparent weight

m is the mass of your body

g is acceleration due to gravity, = 9.8 m/s²

m = w/g

m = 600 / 9.8

m = 61.225 kg

The reading of the scale is now calculated as;

R = m(g-a)

R = 61.225(9.8 - 2.5)

R = 446.94 N

Therefore, the reading of the scale during the acceleration is 446.94 N

4 0

3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago

758 j of heat are added to 0.750 kg of copper. how much does its temperature change?(unit=degrees c) PLEASE HELPPPPPPP MEEEEEE

DiKsa [7]

Answer:

$2.6^{\circ}C$

Explanation:

When a substance is supplied with a certain amount of heat energy, the temperature of the substance increases according to the equation

$Q=mC\Delta T$

where

m is the mass of the substance

Q is the amount of energy supplied

C is the specific heat of the substance

$\Delta T$ is the temperature change

In this problem:

Q = 758 J is the energy supplied

m = 0.750 kg is the mass of the sample

$C=385 J/kg^{\circ}C$ is the specific heat of copper

Re-arranging the equation, we can find the increase in temperature:

$\Delta T=\frac{Q}{mC}=\frac{758}{(0.750)(385)}=2.6^{\circ}C$

5 0

4 years ago