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3 years ago

A boulder rolls off a 100 meter cliff with a horizontal velocity of 2 m/s.

Physics

1 answer:

bonufazy [111]3 years ago

7 0

Explanation:

It is given that,

Height of the cliff is 100 m

Its horizontal velocity is 2 m/s

The attached figure shows the sketch of the cliff and the boulder. It will go in the form of parabola. Its trajectory is parabolic in nature. The horizontal distance is called the range of the parabola.

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(1- name 1 of Faouzia’s songs from her stripped album

Mnenie [13.5K]

Answer:

I thought "Born Without a Heart" was pretty good.

Explanation:

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3 years ago

A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

$0.374 L =x - 0.6 L$

x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

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3 years ago

How does radiation differ from conduction?

Natali5045456 [20]

Answer:

Explanation:

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3 years ago

A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.

V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar, F = 534 N

Co-efficient of kinetic friction, $\mu_k$ = 0.3

Now,

The force against the kinetic friction is given as:

$f = \mu_k N = u_k Mg$

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

$f = 0.3\times108\times9.8$

$f = 317.52N$

Now, the net force on to the metal safe is

$F_{Net}= F-f$

Substituting the values in the equation we get

$F_{Net}= 534N-317.52N$

$F_{Net}= 216.48$

also,

$F_{Net}= M\times$ acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe= $\frac{F_{Net}}{M}$

acceleration of the safe= $\frac{216.48}{108}$

acceleration of the safe= $2.00 m/s^2$

Hence, the acceleration of the metal safe will be 2.00 m/s²

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3 years ago

The energy stored in oil come from the sun. Describe how energy from the sun became stored in oil

nirvana33 [79]

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3 years ago

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