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scoundrel [369]

3 years ago

14

Which of the following is a good example of a contact force? A. Earth revolving around the Sun B. a bridge suspended by cables C

. a ball falling downward a few seconds after being thrown upward D. electrically charged hairs on your head repelling each other and standing up

1 answer:

krek1111 [17]3 years ago

5 0

Answer: a bridge suspended by cables

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A football player runs directly down the field for 45m before turning to the right at an angle of 25 degrees from his original d

Fiesta28 [93]

............................................................................... Hello wonderful person <3

5 0

3 years ago

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

Artemon [7]

Answer:

$\frac{F_1}{F_2} =\frac{1}{2}$

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

$F_1=2k\times d$

case 2:

$F_2=2k\times 2d$

therefore, $\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}$

$\frac{F_1}{F_2} =\frac{1}{2}$

4 0

4 years ago

A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels

katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

$v = \sqrt{\frac{T}{\mu}}$

Here,

v = Velocity

$\mu$ = Linear density (Mass per unit length)

T = Tension

Rearranging to find the Period we have that

$T = v^2 \mu$

$T = v^2 (\frac{m}{L})$

As we know that speed is equivalent to displacement in a unit of time, we will have to

$T = (\frac{L}{t}) ^2(\frac{m}{L})$

$T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})$

$T = 5.54N$

Therefore the tension is 5.54N

8 0

3 years ago

Starburst galaxies:

alexgriva [62]

Answer:

are often associated with a galaxy that is colliding with another galaxy.

Explanation:

A starburst galaxy is a galaxy that undergoes very fast formation of stars. The rate at which stars are born is 100 times more than 3 solar masses per year of the Milky Way. The starburst is stage of the formation of a galaxy. After this stage is complete the stars will have used almost all the gas in it. As the star formation rate is very fast the difference between the age of the stars and the galaxy itself is very less. The star formation is triggered by mergers and tidal interactions between gas-rich galaxies.

6 0

3 years ago

The wheel on an upside-down bicycle moves through 18.0 rad in 5.39 s. What is the wheel’s angular acceleration if its initial an

Elden [556K]

Explanation:

Formula to calculate angular acceleration is as follows.

$\Delta (\theta) = \frac{1}{2} \alpha \Delta t^{2} + \omega_{1} \Delta t$

or, $\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}$

Putting the given values into the above formula as follows.

$\alpha = \frac{2(\Delta (\theta) - \omega_{1} \Delta t)}{\Delta t^{2}}$

= $\frac{2(18.0 rad - 2.5 rad/s \times 5.30 s)}{(5.39)^{2}}$

= 0.326 $rad/s^{2}$

Thus, we can conclude that the wheel’s angular acceleration if its initial angular speed is 2.5 rad/s is 0.326 $rad/s^{2}$ .

5 0

3 years ago