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snow_tiger [21]
3 years ago
15

PLEASE HELP I NEED TO TURN IT IN IN AN HOUR ITLL GIVE YOU POINTS PLS PLEASE

Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

  1. 17.95025
  2. 172.3995

Explanation:

1.

\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{2.995}{0.16685}\:\\\mathrm{is}\:17.95025

2.

\mathrm{Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer}\\\\910\times\:18945=17239950\\\\910\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\0.18945\mathrm{\:has\:}5\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}5\mathrm{\:decimal\:places}\\\\=172.39950\\\\Refine\\=172.3995

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As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i
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<span>As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road is kinetic friction.

We exert force to move the object from rest and in this case, static friction works. But, when the object comes in motion, then kinetic friction works. Here, since the car is driving without slipping means, kinetic friction acts on it. Its also called sliding or dynamic friction.</span>
5 0
3 years ago
If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)
taurus [48]
Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

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Q = 8 400 000 J

In calories,  4.2 J = 1 Calorie
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Q = 200 000 Calories
4 0
3 years ago
A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu
Murljashka [212]

The work done on the ship is 1.5\cdot 10^7 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

F=5.00\cdot 10^3 N (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

\theta=0^{\circ} (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J

Learn more about work:

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8 0
3 years ago
The length of the track itself is 500m. The vehicles (each of mass 15kg) are dragged
gogolik [260]

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

W = Fd

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

W = (200\ N)(500\ m)\\

<u>W = 100000 J = 100 KJ</u>

5 0
3 years ago
A brother and sister are standing next to each other at rest on a surface of frictionless ice. The brother’s mass is exactly twi
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The sister suddenly pushes her brother. As a result, the sister moves with kinetic energy k. The resulting kinetic energy of the brother is k/2.

<h3>What is kinetic energy?</h3>

The ability of an object to do work by virtue of its motion is called the kinetic energy.

A brother and sister are standing next to each other at rest on a surface of frictionless ice. The brother’s mass is exactly twice that of his sister’s.

If the sister's mass is m, the brother's mass is 2m.

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On comparing, the relation between the kinetic energy of brother and sister is k' = k/2

Thus, the resulting kinetic energy of the brother is k/2.

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