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aleksklad [387]
3 years ago
5

Spiral galaxy rotation curves are generally fairly flat out to large distances. Suppose that spiral galaxies did not contain dar

k matter. How would their rotation curves be different?(A) The orbital speeds would fall off sharply with increasing distance from the galactic center.(B) The rotation curve would be a straight, upward sloping diagonal line, like the rotation curve of a merry-go-round.(C) The orbital speeds would rise upward with increasing distance from the galactic center, rather than remaining approximately constant.(D) The rotation curve would look the same with or without the presence of dark matter.
Physics
1 answer:
Artemon [7]3 years ago
8 0

Answer:

A) The orbital speeds would fall off sharply with increasing distance from the galactic center.

Explanation:

The plot of radial distance versus the orbital speed of objects gives us the galaxy rotation curve. The theoretical and practical curves have significant difference. A possible explanation of this difference could be the existence of dark matter.

According to the theoretical calculations the curve should increase sharply and then decrease as the radial distance increases. The theoretical graph does not take dark matter into account. But the plot made by observations shows the plot increasing first then becoming constant as the radial distance increases.

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During a lab investigation, students added four 50 g masses to two boxes and arranged the boxes so that they were motionless on
IceJOKER [234]

When the resultant force is not equal to zero termed an unbalanced force. By procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

<h3>What is an unbalanced force?</h3>

The forces operating on a body are known as unbalanced forces when the resulting force exerted on it is not equal to zero.

Unbalanced forces acting on the body, causing it to modify its state of motion. To further grasp the nature of imbalanced forces.

<h3 />

The following reasons by which we can understand the unbalanced force caused by the box.

Due to these two reasons, books will move up.

By adding another mass to box 2. The box becomes lighter. As the box becomes lighter the gravity force acting on the box will be less due to which the box easily can move up.

By removing the two masses from box 1. Due to which other become heavier other becomes heavier pulling it down causing box 1 one to go up.

Hence by procedures 4 and 5 students observe an unbalanced upward force on Box 1. Hence option 1 is right for the problem.

To learn more about the unbalanced force refer to the link;

brainly.com/question/227461

3 0
2 years ago
A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it
vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

F = ma...(I)

Using formula of friction force

F= \mu m g....(II)

Put the value of F in the equation (II) from equation (I)

ma=\mu mg....(III)

a = \mu g

Put the value in the equation (III)

a=0.24\times9.8

a=2.352\ m/s^2

We need to calculate the distance,

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}2.352\times(3.0)^2

s=10.584\ m\ approx\ 11\ m

Hence, The distance is 11 m.

3 0
3 years ago
Can someone help me with this problem?
emmasim [6.3K]

Answer:

no

Explanation:

u flipped it i cant see

5 0
2 years ago
A wave is traveling at a speed of 15 m/s and it's wavelength is 5 m. Calculate the waves frequency
Anvisha [2.4K]

Answer:

The frequency of this wave is 3\; \rm Hz.

Explanation:

The frequency f of a wave is the number of wavelengths that this wave covers in unit time (typically a second.)

The wave in this question travels at v = 15\; \rm m \cdot s^{-1}. In other words, this wave covers 15\; \rm m in unit time (a second.) How many wavelengths \lambda would that 15\; \rm m\; correspond to?

The question states that the wavelength of this wave is \lambda = 5\; \rm m. Therefore, there would be 15 / 5 = 3 wavelengths in the 15\; \rm m span that this wave covered in the unit time of one second (1\; \rm s.) Hence, the frequency of this wave would be 3\; \rm s^{-1} (three per second,) which is equivalent to 3\; \rm Hz (three Hertzs.)

In general, the frequency f of a wave with speed v and wavelength \lambda would be:

\displaystyle f = \frac{v}{\lambda}.

For the wave in this question:

\begin{aligned}f &= \frac{v}{\lambda} \\ &= \frac{15\; \rm m \cdot s^{-1}}{3\; \rm s} = 3\; \rm s^{-1} = 3\; \rm Hz\end{aligned}.

3 0
3 years ago
A 0.120 kg baseball, thrown with a speed of 40.6 m/s, is hit straight back at the pitcher with a speed of 49.9 m/s.a) What is th
ad-work [718]

Answer:

(A) Impulse will be 1.116 kgm/sec

(B) Force will be equal to 1116 N

Explanation:

We have given mass of the basketball m = 0.120 kg

Initial speed of the ball v_1=40.6m/sec

Final speed of the ball v_2=49.9m/sec

(A) Impulse delivered by the ball is equal to the change in momentum

So impulse will be equal to =m(v_2-v_1)=0.120\times (49.9-40.6)=1.116kgm/sec

So impulse will be 1.116 kgm/sec

(B) Time is given for which force is exerted =10^{-3}sec

We know that impulse is equal to Impluse=force\times time

1.116=force\times10^{-3}

F = 1116 N

6 0
3 years ago
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