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Oksana_A [137]
3 years ago
13

What is the primary force that holds things together on an astronomical scale is________?

Physics
1 answer:
Svetllana [295]3 years ago
5 0
The <span>primary force that holds things together on an astronomical scale is "Gravitational Force"

Hope this helps!</span>
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During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Thepotemich [5.8K]

Answer:

T=6.75s

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on  and the second when the rocket's engines is off. So, we need to know the height (h_1) that the rocket reaches while its engine is on and we need to know the distance (h_2) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\

and the final speed is:

v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2

The sum of these heights will give us the total height, which is known:

h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s

This is the time that its needed in order for the rocket to reach the required altitude.

5 0
3 years ago
A car going 50mph accelerates to pass a truck. five seconds later the car is going 80mph. calculate the acceleration of the car.
gizmo_the_mogwai [7]
Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. <span>Calculation of such is straightforward, if we are given the final velocity, the initial velocity and the total time interval. We can just use the kinematic equations. Fortunately, we are given these values. So, we calculate as follows:
</span>
acceleration = v - v0 / t
acceleration = (80 mph - 50 mph) ( 1 h / 3600) / 5 s
acceleration = 1.67 x 10^-3 m / s^2 

Therefore, the acceleration of the car is 1.67 x 10^-3 m / s.
8 0
3 years ago
Read 2 more answers
The idea that large fibers in the sensory nerves can prevent impulses from reaching the brain and thus prevent the sensation of
allochka39001 [22]

Answer:

Gate control theory of pain

Explanation:

The gate control theory of pain sates that non-painful input closes the nerve gates to painful input, which prevents pain sensation from traveling to the central nervous system (brain).

Therefore, the idea that large fibers in the sensory nerves can prevent impulses from reaching the brain and thus prevent the sensation of pain is part of the gate control theory of pain.

4 0
3 years ago
A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x
DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is \lambda

Total charge, Q = 7 nC = 7\times 10^{- 9} C

At x = 2L,

Electric field, \vec{E_{2L}} = 500N/C

Coulomb constant, K = 8.99\times 10^{9} N.m^{2}/C^{2}

Now, we know that:

\vec{E} = K\frac{Q}{x^{2}}

Also the line charge density:

\lambda = \frac{Q}{L}

Thus

Q = \lambda L

Now, for small element:

d\vec{E} = K\frac{dq}{x^{2}}

d\vec{E} = K\frac{\lambda }{x^{2}}dx

Integrating both the sides from x = L to x = 2L

\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx

\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]

\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}

Similarly,

For the field in between the range 2L< x < 3L:

\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx

\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]

\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}

Now,

If at x = 2L,

\vec{E_{2L}} = 500 N/C

Then at x = 3L:

\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C

4 0
4 years ago
Two train whistles, and , each have a frequency of 392 Hz. is stationary and is moving toward the right (away from ) at a speed
trasher [3.6K]

Answer:

Part a)

f = 371.1 Hz

Part b)

f = 417.7 Hz

Part c)

beat frequency = 46.6 Hz

Explanation:

Part a)

Due to doppler's Effect the frequency of the sound heard by the train which is moving away from the observer is given as

f_1 = f_0\frac{v + v_o}{v + v_s}

f_1 = 392(\frac{340 + 15}{340 + 35})

f_1 = 371.1 Hz

Part b)

Now from the second train which is approaching the person we can say

f_2 = f_0\frac{v - v_o}{v - v_s}

f_2 = 392(\frac{340 - 15}{340 - 35})

f_2 = 417.7 Hz

Part c)

As we know that beat frequency is the difference in the frequency from two sources

f_b = f_2 - f_1

f_b = 417.7 - 371.1 = 46.6 Hz

8 0
3 years ago
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