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Rina8888 [55]
3 years ago
9

A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,00

0 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.
Physics
1 answer:
Savatey [412]3 years ago
7 0

Answer:

The answer is "Option B".

Explanation:

r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\

=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\

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which measurment describes the amount of heat needed to raise the temperature of one gram of a material by one degree celsius?
ella [17]

<u>Answer:</u>

Specific Heat

<u>Explanation:</u>

Specific heat is the measurement which describes the amount of heat needed to raise the temperature of one gram of a material by one degree Celsius.

It is the amount of heat required per unit mass to raise the temperature by one degree Celsius. The relationship between heat and the temperature change is usually expressed as shown below:

Q=cmΔT

where Q = heat added,

c= specific heat,

m=mass; and

ΔT = change in temperature


7 0
4 years ago
A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0
Stella [2.4K]

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\

If we carry out summation of Torques on the pulley we will have to,

F_2*d-F_1*d = I \alpha

Where,

I = Inertia moment

\alpha =Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

I = \frac{1}{2} mr^2

Replacing this equations we have know that

(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})

F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}

F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1

Or

F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}

Replacing our values we have that

F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}

F_1 = 0.974 N

Therefore the tension in the string between the pulley and the cart is  0.974 N

6 0
4 years ago
A circuit is shown in the picture above. The chemical energy in the battery is used to light up the light bulb on the other end.
Ainat [17]
To be honest, the picture is so far above that I can't see it at all.
But reading the information in the question's statement, I'd say
the blank should be filled in so that it says:

<span>   The chemical energy in the battery is used to light up the light bulb
   on the other end. Chemical energy in the battery is transformed into
   electrical energy which runs through the circuit wire. (A)</span>
8 0
4 years ago
Read 2 more answers
A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity
Ivahew [28]

A car with a velocity of 22 m/s is accelerated at a rate of 1.6 m/s^2 for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 m/s^2

We will be using the equation of motion:

v = u + at

\therefore v=22+1.(6.8)

\Rightarrow v=22+10.88

\Rightarrow v=32.88 \ m/s

The final velocity become 32.88 m/s.

To learn more about Attention here:

https://brainly.in/question/10557838

#SPJ4

3 0
2 years ago
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blsea [12.9K]

As we know that Nth harmonic of the string is given by

f = \frac{N}{2L}\sqrt{\frac{T}{m/L}}

now here we will have

m/L = mass density = 2 g/m

m/L = 0.002 kg/m

Length = L = 0.600 m

Tension = T = 50.0 N

now from above formula we have

f = \frac{N}{2(0.600)}\sqrt{\frac{50.0}{0.002}}

f = 131.8N

now for first harmonic N = 1

f_1 = 131.8 Hz

for second harmonic N = 2

f_2 = 263.5 Hz

for third harmonic N = 3

f_3 = 395.3 Hz

8 0
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