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Gala2k [10]
3 years ago
15

If the work done to stretch an ideal spring by 4.0 cm is 6.0 J, what is the spring constant (force constant) of this spring? A)

300 N/m B) 3000 N/m C) 3500 N/m D) 7500 N/m E) 6000 N/m
Physics
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

D) 7500 N/m

Explanation:

Given :

Work done by the spring, W = 6 J

Stretch in the spring, x = 4 cm = 0.04 m

Consider k be the spring constant.

The relation between work done and stretch of the spring is:

W=\frac{1}{2}kx^{2}

Substitute the suitable values in the above equation.

6=\frac{1}{2}k(0.04)^{2}

k=\frac{12}{1.6\times10^{-3} }

k = 7500 N/m

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Astronomers observe two separate solar systems each consisting of a planet orbiting a sun. The two orbits are circular and have
Oxana [17]

Answer:

(D) 3

Explanation:

The angular momentum is given by:

\vec{L}=\vec{r}\ X \ \vec{p}

Thus, the magnitude of the angular momenta of both solar systems are given by:

L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}

where we have taken that both systems has the same radius.

By taking into account that T1=3T2, we have

L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1

but L1=L2=L:

L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3

Hence, the answer is (D) 3

HOPE THIS HELPS!!

3 0
3 years ago
Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would
Butoxors [25]

Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would the composition and sizes of the planets of the inner solar system be different from what we see today is given below

Explanation:

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2.The frost line in the solar nebula lies between Mars and Jupiter. It is the distance where it was cold enough for hydrogen compounds to condense into ices. Frost line: Explain how temperature differences led to the formation of two distinct types of planets.

3. The frost line is the point moving away from the Sun where it is cool enough for hydrogen compounds to freeze. Since the solar nebula was hotter near the center of the disk, hydrogen compounds such as water stayed gaseous in the inner solar system. Outside of the frost line, they froze.

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4 0
2 years ago
9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?
kakasveta [241]

Answer:

Option B (1.51 m)

Explanation:

U 2 can help me by marking as brainliest........

5 0
2 years ago
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during a workout, a football player pushes a blocking dummy a distance of 30 m. while pushing the dummy the same distance a seco
babymother [125]
Power=\frac{Work}{Time}=\frac{Force\times distance}{time}

If he wants to increase power, force must increase and decrease time.
8 0
3 years ago
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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

7 0
3 years ago
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