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3 years ago

There is a bell at the top of a tower that is 45 m high. The bell weighs 190 kg. The bell has

Physics

1 answer:

natali 33 [55]3 years ago

6 0

Answer: The bell has 8550 J energy.

given, There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N

i.e., bell is located at the top of tower, h = 45m

weight of the bell, F = 190 N

workdone by the gravitational force = F.hcos180°

[ gravitational force (i.e., weight ) acting downward while body is located 45m above the ground. so, angle between force and h = 180° ]

workdone by the gravitational force = 190 × 45 × (-1)

= -8550 J

we know, potential energy = negative of workdone

= -(-8550 J) = 8550 J

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4 years ago

Consider projectile thrown horizontally at 50 m/s from height of 19.6 meters. The projectile will take ______________ time to hi

aleksley [76]

Answer:

C)The Same

Explanation:

Kinematics equation:

$y=v_{oy}*t+1/2*g*t^2$

for both cases the initial velocity in the axis Y is the same, equal a zero.

So the relation between the height ant temps is the same for both cases (the horizontal velocity does not play a role)

C)The Same

3 0

3 years ago

The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le

Vinil7 [7]

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

3 0

4 years ago

A helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m

Advocard [28]

Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

We have to find the speed of helicopter ascending at the moment when the passenger let go of the wallet.

$v^2-u^2=2gh$

Where $g=9.8 m/s^2$

Substitute the values

$0-u^2=2(-160)\times 9.8$

$u^2=3136$

$u=\sqrt{3136}=56m/s$

Option (E) is true

8 0

4 years ago

A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit

elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

= 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

$T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$ second

Squaring the terms and solving it for 'g'

g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$ m/s²

Substituting the values in the above equation

g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$

g = 4.84 m/s²

Therefore, the weight

w = m x g newton

= 5832 Kg x 4.84 m/s²

= 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N

8 0

4 years ago