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2 years ago

Please help me guys please

Physics

1 answer:

leonid [27]2 years ago

8 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Solution is in attachment ~

I hope that you got what you were looking for, and if there's different data then go through the same procedure, using same formula with different values and you will get your answer ~

$\mathrm{✌TeeNForeveR✌}$

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2 years ago

How far will a 600 kg boat travel in 10 s if there is a constant 900 N force on it and it starts from rest?

natta225 [31]

Answer:

here

Explanation:

There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. ... a 3.25-kg object rightward with a constant acceleration of 1.20 m/s/s if the force of ... of 33.8 kg, how far (in meters) will it move in 1.31 seconds, starting from rest?

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2 years ago

The mass of a star is 1.61·1031 kg and its angular velocity is 1.60E-7 rad/s. Find its new angular velocity if the diameter sudd

Tcecarenko [31]

Answer:

ω₂ = 1.9025 x 10⁻⁶ rad/s

Explanation:

given,

mass of star = 1.61 x 10³¹ kg

angular velocity = 1.60 x 10⁻⁷ rad/s

diameter suddenly shrinks = 0.29 x present size

r₂ = 0.29 r₁

using conservation of angular momentum

I₁ ω₁ = I₂ ω₂

$(\dfrac{2}{5}mr_1^2)\omega_1=(\dfrac{2}{5}mr2^2)\omega_2$

$r_1^2\times \omega_1=r_2^2\times \omega_2$

$r_1^2\times 1.60\times 10^{-7}=(0.29r_1)^2\times \omega_2$

$1.60\times 10^{-7}=0.0841\times \omega_2$

$\omega_2=\dfrac{1.60\times 10^{-7}}{0.0841}$

ω₂ = 1.9025 x 10⁻⁶ rad/s

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3 years ago

A certain wave has a wavelength of 35 meters and a frequency of 4.0hz. What is the speed of the wave

defon

The speed of the wave is 140

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2 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

2 years ago

Please help me guys please​

Please help me guys please