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3 years ago

A rock at rest falls off a tall cliff and hits the valley below after 3.5s. What is the rocks velocity as it hits the ground

Physics

2 answers:

pogonyaev3 years ago

7 0

T = 3.5 secs

Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s

Paraphin [41]3 years ago

5 0

V = u + at
u=0(rest) ; a=gravitational acc(g)=10
v = 0 + 10(3.5) = 35m/s

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Answer:

10cm^3

Explanation:

Given data

Mass of steel = 80grams

density of steel= 8 g/cm^3

We know that the formula for density is given as

density= mass/volume

make volume subject of formula

volume= mass/density

volume=80/8

volume= 10cm^3

Hence, the volume 10cm^3

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3 years ago

Three forces act on an object. Two of the forces have the magnitudes 56 N and 27 N, and make angles 53° and 156°, respectively,

Ludmilka [50]

Answer:

Explanation:

Given

Two forces $F_1$ and $F_2$ at an angle of $\theta _1$ $\theta _2$

$F_1=56\ N$

$F_2=27\ N$

$\theta _1=53^{\circ}$

$\theta _2=156^{\circ}$

As resultant force is zero therefore horizontal component as well as vertical component of force is zero

$\sum F_x=F_1\cos \theta _1+F_2\cos \theta _2+F_3\cos \theta _3$

$\sum F_x=56\cos 53+27\cos 156+F_3\cos \theta$

$F_3\cos \theta_3=-9.035\ N----1$

$\sum F_y=F_1\sin \theta _1+F_2\sin \theta _2+F_3\sin \theta _3$

$F_3\sin \theta _3=55.704\ N----2$

squaring and adding 1 and 2

$F_3^2(\cos ^2\theta _3+\sin^2\theta _3)=86.583+3102.93$

$F_3=56.47\ N$

Divide 1 and 2 to get $\theta _3$

$\frac{\sin \theta_3 }{\cos \theta_3 }=\frac{-55.704}{9.305}$

$\tan \theta_3=-6.16$

$\theta _3=180-80.78$

$\theta _3=99.22^{\circ}$

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3 years ago

The idea that Earth’s lithosphere is divided into large, moving sections is called the _____. biosphere theory global positionin

victus00 [196]

The answer to this is the plate tectonics theory.

The earth's lithosphere is divided into large, moving sections which are called the tectonics plates.

Hope this helped :)

Have a great day

5 0

3 years ago

Read 2 more answers

A positive charge is moved from point a to point b along an equipotential surface. How much work is performed or required in mov

snow_lady [41]

Answer:

The work done is zero

Explanation:

No work is performed or required in moving the positive charge from point A to point B

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3 years ago

What is the electric flux ΦΦPhi through each of the six faces of the cube?

BlackZzzverrR [31]

Flux through each of the six faces of the cube: $\frac{q}{6\epsilon_0}$

Explanation:

In this problem, a charge $q$ is placed at the center of the cube.

We can apply Gauss Law, which states that the flux through the surface of the cube is equal to the charge contained within the cube divided by the vacuum permittivity; mathematically:

$\Phi_T=\frac{q}{\epsilon_0}$

where

$q$ is the charge

$\epsilon_0$ is the vacuum permittivity

Here we want to find the flux through each of the six faces of the cube.

By simmetry, we can say that the 6 faces are identical: therefore, the flux through each of them must be the same. This means that the flux through each faces is 1/6 of the total flux through the total surface, therefore:

$\Phi = \frac{1}{6}\Phi = \frac{q}{6\epsilon_0}$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago