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S_A_V [24]
3 years ago
8

Each of the colors of light in the visible light spectrum has...

Physics
1 answer:
lara31 [8.8K]3 years ago
7 0
The answer is d that might help you
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Consider the table below. Which of the following data sets depict an accelerating object? Mark all that apply.
Gre4nikov [31]

Answer:

A and B

Explanation:

The data sets that depict an accelerating object is Data Set A & Data Set B.

The both data sets show that the body is accelerating. Also, they show that the body started from rest (0m/s) at a 0sec.

Data Set A shows a non-constant acceleration which has changing amount of velocity with change in time. While Data Set B shows a constant acceleration which has constant amount of velocity with change in time.

7 0
3 years ago
A simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top
Trava [24]

Answer:

He will complete the race in total time of T = 10 s

Explanation:

Total distance moved by the sprinter in 2.14 s is given as

s = \frac{(v_{in} + v_{f})}{2} time

s = \frac{(0 + 11.2)}{2} (2.14)

s = 11.98 m

now the distance remaining to move

d = 100 - 11.98 = 88 m

now he will move with uniform maximum speed for the remaining distance

so we will have

time = \frac{d}{v}

time = \frac{88}{11.2} = 7.86 s

so the total time to complete the race is given as

T = 7.86 + 2.14 = 10 s

6 0
3 years ago
Can somebody help please !<br><br> a. -8.3 m/s<br> b.-4.2 m/s<br> c.-0.12 m/s<br> d. 0 m/s
Gnom [1K]
The answer is a
the \: answer \: is \: a.
4 0
4 years ago
Halleys comet has period of 75.3 years. Using Kepler’s third law, find it’s semimajor axis expressed in astronomical units?
natta225 [31]

Answer: 17.83 AU

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  </em>

T^{2}\propto a^{3}  (1)

Talking in general, this law states a relation between the <u>orbital period</u> T of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u> a of its orbit.

However, if T is measured in <u>years</u>, and a is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:

T^{2}=a^{3}  (2)

This means that now both sides of the equation are equal.

Knowing T=75.3years and isolating a from (2):

a=\sqrt[3]{T^{2}}=T^{2/3}  (3)

a=(75.3years)^{2/3}  (4)

Finally:

a=17.83AU  (5)

4 0
3 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
4 years ago
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