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4 years ago

How does activity on the Sun affect natural phenomena on Earth?

Physics

1 answer:

Nana76 [90]4 years ago

5 0

Answer and Explanation:

The Sun is the main source of energy on the earth if there will be no availability of Sun energy then life is impossible om the earth besides this the Sun warms our planet. The heating of ocean and atmosphere is mainly sue to Sun energy .Sun has also a great impact on the weather we can say that Sun is weather deciding on the earth our climate is totally dependent on the how much energy we got in form of radiation from earth.

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Lol somebody tell me the answer ill also give 5 stars

sdas [7]

Hi C is the Answer I believe

3 0

3 years ago

Consider an electrical transformer that has 10 loops on its primary coil and 20 loops on its secondary coil. What is the current

rusak2 [61]

Answer:

C. 2.5 A

Explanation:

Transformer: A transformer is an electromechanical device that is used to change the voltage of an alternating current.

The current and the number of loops in a transformer is related as shown below

Ns/Np = Ip/Is........................... Equation 1

Where Ns = Secondary loop, Np = primary loop, Ip = primary current, Is = secondary current.

Making Is the subject of the equation

Is = NpIp/Ns........................ Equation 2

Given: Np = 10 loops, Ns = 20 loops, Ip = 5.0 A.

Substitute into equation 2

Is = (10×5.0)/20

Is = 50/20

Is = 2.5 A.

Hence the current in the primary coil = 2.5 A.

The right option is C. 2.5 A

6 0

3 years ago

Bone has a Young's modulus of about

saveliy_v [14]

Answer: 4.74 mm

Explanation:

We can solve this problem with the following equation:

$Y=\frac{stress}{strain}$ (1)

Where:

$Y=1.8(10)^{10} Pa$ is the Young modulus for femur

$stress=\frac{F}{A}=1.58(10)^{8} Pa$ is the stress (force $F$ applied per unit of transversal area $A$ ) on the femur

$strain=\frac{\Delta l}{l_{o}}$

Being:

$\Delta l$ the compression the femur can withstand before breaking

$l_{o}=0.54 m$ is the length of the femur without compression

Writing the data in equation (1):

$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l_{o}}}$ (2)

$1.8(10)^{10} Pa=\frac{1.58(10)^{8} Pa}{\frac{\Delta l}{0.54 m}}$ (3)

Isolating $\Delta l$ :

$\Delta l=\frac{(1.58(10)^{8} Pa)(0.54 m)}{1.8(10)^{10} Pa}$ (4)

$\Delta l=0.00474 m$ (5) This is the compression in meters

Converting this result to millimeters:

$\Delta l=0.00474 m \frac{1000 mm}{1 m}=4.74 mm$

4 0

3 years ago

All BUT one factor will increase the work output of a machine and decrease the amount of thermal energy created and released. Th

Serggg [28]

I would say C because extra friction would make the machine work more and therefore create more heat.

6 0

4 years ago

Read 2 more answers

A student stretches a spring, attaches a 1.20 kg mass to it, and releases the mass from rest on a frictionless surface. The resu

Sphinxa [80]

Answer:

the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

Explanation:

Given that;

mass of the object m = 1.20 kg

period of oscillation = 0.750 s

Amplitude ( A/x) = 15.0 cm = 0.15 m

now;

a) Determine the oscillation frequency;

oscillation frequency f = 1/T

we substitute

f = 1 / 0.750 s

f = 1.33 Hz

Therefore, the oscillation frequency is 1.33 Hz

b) Determine the spring constant;

we solve for spring constant from the following expression;

T = 2π√(m/k)

k = 4π²m / T²

so we substitute

k = (4π² × 1.20) / (0.750)²

k = 47.3741 / 0.5625

k = 84.22 N/m

Therefore, the spring constant is 84.22 N/m

c) determine the speed of the mass when it is halfway to the equilibrium position

So, at equilibrium, the energy is equal to K.E

such that;

1/2mv² = 1/2kx²

mv² = kx²

v² = kx² / m

v = √( kx²/m)

we substitute

v = √( 84.22×(0.15 m)²/ 1.2 )

v = √( 1.89495 / 1.2 )

v = √ 1.579125

v = 1.26 m/s

Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

3 0

3 years ago