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3 years ago

13

HELP PLEASE : ) IT'S SCIENCE

1 answer:

nekit [7.7K]3 years ago

3 0

Answer:

1. A and C are balanced forces as all the forces get cancelled by the opposite and equal force.

2. B and D as in B the Net force is 5N to the right and in D the Net force is 20N to the right.

3. This would be all the unbalance forces which are B and D

4. B would be moving 5N to the right and D would move 20N to the right.

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This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa

Montano1993 [528]

Answer:

$2.77287\times 10^{15}\ m$

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

$\nu$ = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

$P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s$

Photon intensity is given by

$i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m$

The distance is $2.77287\times 10^{15}\ m$

3 0

3 years ago

A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel

lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

$\epsilon = - d\frac{\phi_B}{dt}$

$\epsilon = -NA\frac{dB}{dt}$

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

$\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}$

$\epsilon = 0.212V$

$\epsilon = 0.21V$

Therefore the correct answer is A.

6 0

3 years ago

Which of the following is an electromagnetic wave?

morpeh [17]

Your answer will be Radio Waves .

That seems to be the only to make sense. Hope that helps u

5 0

3 years ago

Read 2 more answers

Which part of the microscope is the circular area on the stage that light passes through?

patriot [66]

Answer: The part of the microscope that is the circular area is the APERTURE

I hope this helped!

6 0

1 year ago

A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can

GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

$v$ = speed at which jumper jumps at all time

initial distance jumped is given as

$R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}$

final distance jumped is given as

$R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}$

Dividing final distance by initial distance

$\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}$

$\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}$

$R_{2} =7.38$

distance lost is given as

d = $R_{1} - R_{2}$

d = 7.4 - 7.38

d = 0.02 m

8 0

3 years ago