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Elodia [21]
3 years ago
15

REAL ANSWERS ONLY PLS

Physics
1 answer:
just olya [345]3 years ago
7 0

Answer:

The statement of the student is correct.

Since B attained a higher velocity in a short amount of time, that is it accelerated faster(having a larger slope).

Slope = dy/dx

That is, <u>Velocity</u>

Time

which is acceleration.

That's my guess.

Hope it's right.

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A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harm
Marina86 [1]

Answer:

v(0)=2m/s

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

v(t)=\omega  *A*cos(\omega t + \phi)

Where:

\omega=Angular\hspace{3}frequency\\A=Amplitude\\\phi=Initial\hspace{3}phase

Express the amplitude in meters:

10cm*\frac{1m}{100cm} =0.1m

The angular frequency can be found using the next equation:

\omega=\sqrt{\frac{k}{m} }

Using the data provided:

\omega=\sqrt{\frac{400}{1} } =20

At the equilibrium position:

\phi=0

v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s

6 0
3 years ago
Gordon throws a baseball into the air. It rises, stops when it reaches its greatest height, and then falls back down to the grou
Arte-miy333 [17]

Answer:

A) When the baseball is rising

Explanation:

When an object like the baseball is thrown into the air, the kinetic energy  is gradually converted into potential energy. When the baseball rises to its maximum height, the kinetic energy becomes zero and the kinetic energy is fully converted into potential energy.

7 0
3 years ago
Hubble law how did astronomers gather the information needed to establish the law?
Korvikt [17]
They measured the wavelength of light emitted by stars using spectrometers and found it was being redshifted.
This implied the stars were moving away aka the space between the scientists and the star was expanding
7 0
3 years ago
6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t
Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of  12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

y = u sin \theta - \dfrac{1}{2}gt^2................(2)

from equation(1) and (2)

y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}..........{3}

12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}

\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29

\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2

u = \sqrt{10959.34}

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0
3 years ago
A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction
Lady_Fox [76]

Answer:

The speed of this light and wavelength in a liquid are 2.04\times10^{8}\ m/s and 442 nm.

Explanation:

Given that,

Wavelength = 650 nm

Index refraction = 1.47

(a). We need to calculate the speed

Using formula of speed

n = \dfrac{c}{v}

Where, n = refraction index

c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

1.47=\dfrac{3\times10^{8}}{v}

v=\dfrac{3\times10^{8}}{1.47}

v= 2.04\times10^{8}\ m/s

(b). We need to calculate the wavelength

Using formula of wavelength

n=\dfrac{\lambda_{0}}{\lambda}

\lambda=\dfrac{\lambda_{0}}{n}

Where, \lambda_{0} = wavelength in vacuum

\lambda = wavelength in medium

Put the value into the formula

\lambda=\dfrac{650\times10^{-9}}{1.47}

\lambda=442\times10^{-9}\ m

Hence, The speed of this light and wavelength in a liquid are 2.04\times10^{8}\ m/s and 442 nm.

3 0
3 years ago
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