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3 years ago

The sun generates energy through what?

Physics

2 answers:

jarptica [38.1K]3 years ago

6 0

Option a
nuclear fusion

OlgaM077 [116]3 years ago

4 0

The correct answer is A. Nuclear fusion

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Solving equation for the quantity indicated

Slav-nsk [51]

T = (vf -vo)/a. .......

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2 years ago

Read 2 more answers

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha

jasenka [17]

Explanation:

It is given that,

Mass of an object, $m=4\times 10^5\ kg$

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

Where

k is the force constant

$k=\dfrac{4\pi ^2 m}{T^2}$

$k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}$

$k=2.74\times 10^6\ N/m$

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

$U=\dfrac{1}{2}kx^2$

$U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2$

$U=6.63\times 10^6\ J$

Hence, this is the required solution.

7 0

3 years ago

Why do these two magnets attract each other?

Liono4ka [1.6K]

Answer:

Explanation:

When two magnets are brought near each other, like poles repel; opposite poles attract. We like to represent this force effect of a magnet on iron-like objects with a concept called magnetic field. The concept of field can be best understood if we remember the gravitational force of Earth on object near it.

5 0

2 years ago

Read 2 more answers

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a

ValentinkaMS [17]

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

$area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v$

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

$E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m$

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

$8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}$

from the expression for the capacitance

$C=eA/d$

if we substitute values we arrive at

$C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF$

d. To calculate the charge on each plate, we use the formula below

$Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC$

8 0

3 years ago

Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to b

weeeeeb [17]

Answer:

4.408 m/s, 4.102 m/s, 4.026 m/s

Explanation:

The question is incomplete. The text of the original question states:

A race car moves such that its position fits the relationship

$x=(4.0 m/s)t + (0.85 m/s^3) t^3$

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.

We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

$v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2$

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

$v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s$

5 0

2 years ago