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Sergeu [11.5K]
3 years ago
10

A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.

Physics
1 answer:
Alex Ar [27]3 years ago
3 0

Good.  You can do some very interesting experiments with that equipment.

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Compare these two waves.
Anastaziya [24]

Answer:

id 272 966 3863

Pas 161005

5 0
2 years ago
The height of the Washington Monument is measured to be 170 m on a day when its temperature is 35.0°C. What will the change in i
Alecsey [184]

Answer:

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m

Explanation:

Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α

So, let us take the average value

(5.5+14.1)/2 = 9.8×10⁻⁶ /K

Change in temperature = 35-(-18) = 53 K = ΔT

Original length = 170 m = L

Linear thermal expansion

\frac{\Delta L}{L} = \alpha\Delta T\\\Rightarrow \Delta L=\frac{\alpha\Delta T}{L}\\\Rightarrow \Delta L=9.8\times 10^{-6}\times 53\times 170

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)

4 0
3 years ago
Suppose two objects are gravitationally attracted to each other with some force F. If the mass of object 1 is multiplied by a fa
RoseWind [281]

If F = Gm₁m₂/d², and we change m₁ to 5m₁ and m₂ to 2m₂, then the new magnitude of the gravitational force is

F' = G (5m₁) (2m₂) / d²

F' = 10  Gm₁m₂ / d²

but this is really just F' = 10F. So J is the correct choice.

8 0
2 years ago
Read 2 more answers
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
Amy counts the wave crest traveling down a stretched string five wave crests pass Amy in two seconds what is the frequency of th
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8 0
3 years ago
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