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3 years ago

a tiger leaps horizontally from a 5.7m high rock with a speed of 4.1m/s. How far from the base of the rock will she land

Physics

1 answer:

shepuryov [24]3 years ago

6 0

Looks like a projectiles problem, with the tiger as the projectile. See enclosed for guidance.For more information/help, please ask.ps there's a poem which contains the lines "Tyger tyger burning bright, in the forest of the night; what immortal hand or eye; could frame thy fearful symmetry ?"

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Which person is weightless?

ahrayia [7]

Answer: Option (b) is the correct answer.

Explanation:

The force of gravity acting on an object helps in determining the weight of an object. But a place where there will be no gravity or have zero gravitational pull then it means the person will be weightless.

For example, force of gravity on moon is zero which means any object or person on moon will be weightless.

On the other hand, when a child is in the air as she plays on a trampoline then it means gravitational pull form the earth is acting on it. So, it will definitely has some weight.

Similarly, a scuba diver exploring a deep-sea wreck is under the ground where there will be force of gravity. Hence, it will also have some weight.

Thus, we can conclude that an astronaut on the Moon is the person who is weightless.

7 0

2 years ago

One factor that could account for a drop in the water table is:

Shtirlitz [24]

Answer:

D. Less rain and snow.

Explanation:

A factor that can be a account for a drop in the water table is, less rain and snow. to topography, water tables is influenced by lot of factors, including the geology, weather, ground cover.

4 0

2 years ago

A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

Luden [163]

Answer:

$\rho _{liquid}=1995.07kg/m^{3}$

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have $T_{3}+B = W_{rock}$

$T_{3}+\rho _{Liquid}V_{rock}g$ = $W_{rock}.....(\alpha)$

When the rock is suspended in air for equilibrium we have

$T_{1}=W_{rock}....(\beta)$

When the rock is suspended in water for equilibrium we have

$T_{2}$ + $\rho _{water}V_{rock}g$ = $W_{rock}.....(\gamma)$

Using the given values of tension and solving α,β,γ simultaneously for $\rho _{Liquid}$ we get

$W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\$

Solving for density of liquid we get

$\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000$

$\rho _{liquid}=1995.07kg/m^{3}$

5 0

2 years ago

-1

Rom4ik [11]

Answer:

9800 N

Explanation:

applying,

W = Fd.................. Equation 1

Where W = work done by the engine, F = Force exterted to lift the beam, d = distance.

make F the subject of the equation

F = W/d............... Equation 2

From the question,

Given: W = 1421000 J, d = 145 meters.

Substitute these values into equation 2

F = 1421000/145

F = 9800 N

5 0

2 years ago

3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to th

Fed [463]

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

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nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn

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2 years ago