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Alla [95]
3 years ago
13

A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

oscillations having a period of 1.60 s. Find the force constant of the spring.
Physics
1 answer:
mezya [45]3 years ago
5 0

Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.

N/m

Explanation:

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The Hydrogen bomb is a thermonuclear bomb where a fission reaction is used to trigger a fusion reaction.

Answer: Option A

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8 0
3 years ago
Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of them is moved sideways by 4.0 mm
loris [4]

Answer:

4.4721m

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a^+b^2=c^2\\4^2+2^2=d^2\\\sqrt{20}\ mm=d

There are antinodes 1/4,1/2 and 3/4 of the distance between speakers.

The greatest antinode is 3/4-1/4=1/2

#Distance between consecutive antinodes is:

0.5\times \sqrt{20}=\lambda/2\\\\\lambda=4.4721m

Hence, the maximum possible wavelength of the sound waves is 4.4721m

6 0
3 years ago
A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. Th
Hitman42 [59]

Answer:

Explanation:

Given

inclination \theta =30^{\circ}

initial speed u=20\ m/s

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

t_1=\frac{2u\sin \theta }{g}

t_1=\frac{2\times 20\times \sin 30}{10}

t_1=2\ s

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

h=u_yt+\frac{1}{2}a_yt^2

where, h=height

u_y=vertical velocity

a_y=vertical acceleration

t_0=time

45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2

t_0^2=\frac{70}{9.8}

t_0=2.64\ s

thus total time time required is t=t_0+t_1=2.64+2=4.64\ s

vertical velocity just before hitting

v_y=\sqrt{u_y^2+2\times a_y\times s}

v_y=\sqrt{10^2+2\times 10\times 45}

v_y=\sqrt{1000}=31.622\ m/s

Horizontal velocity v_x=u\cos 30=17.32\ m/s

Net velocity Just before hitting =\sqrt{v_x^2+v_y^2}

=\sqrt{(17.32)^2+(31.62)^2}

=\sqrt{1299.82}=36.05\ m/s

                 

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Explanation:

Hope it helps you................

Plz mark as brainlist answer

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