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2 years ago

Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal

Physics

2 answers:

Serhud [2]2 years ago

6 0

Mass doesn't matter on amount of work done .We can calculate amount of work done through Force and Displacement

Force=87N
Displacement=4.5m

$\boxed{\sf W=Fd}$

W denotes to work done
F denotes to force
d denotes to displacement

$\\ \sf\longmapsto Work\:Done=87(4.5)$

$\\ \sf\longmapsto Work\:done=391.5J$

elena-s [515]2 years ago

5 0

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

W = F║d
where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

W = (87 N)(4.5 m)
W = 391.5 J

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

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What is the power through a device with a resistance of 100 ohms if a current of 8 A is running through it?

Kazeer [188]

Answer:

6400 W (or) 6.4 KW

Explanation:

Formula we use,

→ P = I²R

Let's solve for the power of device,

→ P = I²R

→ P = (8)² × 100

→ P = 64 × 100

→ [ P = 6400 W ]

Hence, the power is 6400 W.

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2 years ago

A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the ci

Anettt [7]

Answer:

The radius is $r = 3.1905 \ m$

Explanation:

From the question we are told that

The distance beneath the liquid is $d = 2.70 \ m$

The refractive index of the liquid is $n_i = 1.31$

Now the critical value is mathematically represented as

$\theta = sin ^{-1} [\frac{1}{n_i} ]$

substituting values

$\theta = sin ^{-1} [\frac{1}{131} ]$

$\theta = 49.76^o$

Using SOHCAHTOA rule we have that

$tan \theta = \frac{ r}{d}$

=> $r = d * tan \theta$

substituting values

$r = 2.7 * tan (49.76)$

$r = 3.1905 \ m$

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3 years ago

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3 years ago

Read 2 more answers

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

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3 years ago

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AlladinOne [14]

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3 years ago