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valentinak56 [21]
2 years ago
14

A 50.0 kg object is moving at 18.2 m/s when a 200 N force

Physics
1 answer:
gayaneshka [121]2 years ago
7 0

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity  = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a  = force ÷ mass

a = \frac{-200}{50}   =  -4 m/s²

we get here now required time that is

required time = \frac{V_{(final)} - V_{(initial)}}{a}     ...............1

put here value

required time = \frac{12.6-18.2}{-4}  

so distance will be

distance = \frac{V_{(final)}^2 - V_{(initial)}^2}{2a}    ........2

distance = \frac{12.6}^2 -{18.2}^2}{2\times (-4)}  

distance = 21.56 m

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Determine the torque applied to the shaft of a car that transmits 225 hp
Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu

8 0
2 years ago
What two processes practiced by scientists increase the likelihood of a successful outcome in science?
Amanda [17]

Answer:

ULTIMATE CORRECT ANSWER

collaboration and communication

Explanation:

5 0
2 years ago
A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm). at the top of the circle the spee
gayaneshka [121]

Answer:

Tension in the string at this position: 3.1 N.

Explanation:

Convert the radius of the circle to meters:

r = 80\;\text{cm} = 0.80\;\text{m}.

What's the net force on the object?

The object is in a circular motion. As a result,

\displaystyle \Sigma F = \frac{m\cdot v^{2}}{r},

where

  • \Sigma F is the net force on the object,
  • m is the mass of the object,
  • v is the velocity of the object, and
  • r is the radius of the circular motion.

For this object,

\displaystyle \Sigma F = \frac{0.20\times {4.5}^{2}}{0.80} = 5.0625\;\text{N}.

The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.

What are the forces on this object?

There are two forces on the object at this moment:

  • Weight, W, which points downwards. W = m\cdot g = 0.20\times 9.81 = 1.962\;\text{N}.
  • Tension, T, which also points downwards. The size of the tension force needs to be found.

What's the size of the tension force?

Gravity and tension points in the same direction. The size of their resultant force is the sum of the two forces. In other words,

\Sigma F = T + W.

T = \Sigma F - W = 5.0625 - 1.962 = 3.1.

All three values in this question are given with two sig. fig. Round the value of T to the same number of significant figures.

4 0
2 years ago
A constant eastward horizontal force of 70 newtons is applied to a 20 -kilogram crate moving toward the east on a level floor. I
matrenka [14]

Answer:

a = 3 m/s²

Explanation:

given,

mass of crate = 20 Kg

horizontal force on crate = 70 N

frictional force on the crate = 10 N

acceleration of crate = ?

now, calculating net force acting on the crate.

 F = horizontal force - frictional force

 F = 70 - 10

 F = 60 N

net force on the crate is equal to 60 N.

We also know that

F = m a

60 = 20 x a

a = 3 m/s²

Hence, the acceleration of the crate is equal to 3 m/s²

7 0
2 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
2 years ago
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