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3 years ago

What is the kinetic energy of a toy truck with a mass of 0.75 kg and a velocity of 4 m/s? (Formula: )

Physics

2 answers:

sweet-ann [11.9K]3 years ago

6 0

Kinetic energy is mass times velocity square mv^2 (0.75*4^2=12)

sukhopar [10]3 years ago

5 0

Answer: 6 J .

Explanation:

What is the kinetic energy of a toy truck with a mass of 0.75 kg and a velocity of 4 m/s? (Formula: )

3 J

6 J

12 J

24 J

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An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and

Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

8 0

3 years ago

If the person drops box from 3.8 m how much energy is transferred from potential energy to kinetic energy

kotykmax [81]

Answer:

Kinetic energy

When work is done the energy is transferred from one type to another. This transferred energy may appear as kinetic energy.

For example, when you pedal your bicycle so that its speed increases, you are doing work to transfer chemical energy from your muscles to the kinetic energy of the bicycle.

Kinetic energy is the energy an object possesses by virtue of its movement. The amount of kinetic energy possessed by a moving object depends on the mass of the object and its speed. The greater the mass and the speed of the object the greater its kinetic energy.

The kinetic energy Ek of an object of mass m at a speed v is given by the relationship

{E_k} = \frac{1}{2}m{v^2}

m is the mass of the object in kilograms ( kg) and v is the speed of the object in metres per second ( m\,s^{-1}).

Explanation:

When work is done on an object it may also lead to energy being transferred to the object in the form of gravitational potential energy of the object.

Gravitational potential energy is the energy an object has by virtue of its position above the surface of the Earth. When an object is lifted, work is done. When work is done in raising the height of an object, energy is transferred as a gain in the gravitational potential energy of the object.

For example, suppose you lift a suitcase of mass m through a height h. The weight W of the suit case is a downward force of size mg. In lifting the suitcase, you would have to pull upwards on it with a force equal in size to its weight, mg.

Two suitcases. One has a green force arrow pointing up labelled F and a purple force arrow pointing down labelled 'Weight = mg'. The other case is raised by a height labelled h.

Suitcases with forces and height labelled

When this force (equal to the weight mg, but upwards) is applied to the suitcase over the distance h:

Work\,done=force\,\times\,distance\,upwards=mg\,\times\,h

This energy is transferred to potential energy when raising the object through a known height.

Energy = mass \times gravitational\,field\,strength \times height

E = m \times g \times h

This is the relationship used to calculate gravitational potential energy.

{E_p} = mgh

where m is the mass of the object in kilograms (kg), g is the gravitational field strength, (for positions near the surface of the Earth g = 9∙8 newtons per kilogram ( N kg ^{-1} and h is the height above the surface of the Earth in metres ( m).

8 0

3 years ago

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu

galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

$F=I \times B \times L \times sin(\theta)$

So from data

$F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N$

Now sub parts

(a)

$\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N$

(b)

$\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N$

(c)

$\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N$

3 0

3 years ago

Determine la fuerza aplicada sobre un émbolo en una prensa hidráulica si se quiere cargar una lanza de 2970kg; toma en cuenta qu

Artist 52 [7]

I’m not sure sorry I really wish I could help

4 0

2 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago