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3 years ago

If a child applies a 3 N force for 2 m in the same direction the car is already moving, how much work is

Physics

2 answers:

horsena [70]3 years ago

7 0

Answer:6

Explanation:

REY [17]3 years ago

6 0

Answer:

6, this could be wrong. (but I believe it is correct)

Explanation:

w=Fs

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PLEASE HELP ASAP!!!

I am Lyosha [343]

Answer:

B - Velocity

Explanation:

Velocity definition: “The speed of something in a given direction.”

6 0

2 years ago

An object of volume 0.0882 m3 is

vfiekz [6]

The density of the fluid is 776.3 $m^{-3}$

<u>Explanation:</u>

Buoyant force is the upward pushing force whenever an object is trying to get immersed in fluid. So this is the force given by the fluid on the object which is trying to get immersed. The buoyant force is found to be directly proportional to the product of density of the object, volume of the object. And here the acceleration due to gravity will be acting as proportionality constant.

$Buoyant force = Density \times Volume \times Acceleration$

As, buoyant force is given as 671 N and volume is 0.0882 $m^{3}$ and acceleration is known as 9.8 m/ $s^{2}$ . Then density is

$\text { Density }=\frac{\text { Buoyant force }}{\text {Volume } \times \text {Acceleration}}$

Thus,

$\text { Density }=\frac{671}{0.0882 \times 9.8}=\frac{671}{0.86436}=776.296 \mathrm{kg} / \mathrm{m}^{3}$

Density is 776.3 kg $m^{-3}$ .

7 0

3 years ago

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

3 years ago

What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?

goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0

3 years ago

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measur

Over [174]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v $_r$ = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

v $_r$ = ( u + u ) / ( 1 + ( u×u / c²) )

v $_r$ = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )

we substitute

v $_r$ = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )

v $_r$ = 1.904c / ( 1 + 0.906304 )

v $_r$ = 1.904c / 1.906304

v $_r$ = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

5 0

3 years ago