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3 years ago

What is the net force on a 4000 kg car that doubles its speed from 15 m/s to 30 m/s over 10 seconds

Physics

1 answer:

kozerog [31]3 years ago

3 0

The net force of the car <u>is 6000 N.</u>

Why?

To calculate the net force of the car during the given time, we first need to calculate its acceleration, and then, use Newton's Second Law equation to calculate the net force.

We can calculate the acceleration of the car during the given tieme using the following equation:

$a=\frac{v-v_{o}}{time}$

Substituting the given information, we have:

$a=\frac{30\frac{m}{s} -15\frac{m}{s} }{10s}=\frac{15\frac{m}{s} }{10s}=1.5\frac{m}{s^{2} }$

Now that we have calculated the acceleration, we can calculate the net force:

$Force=mass*acceleration\\\\Force=4000kg*1.5\frac{m}{s^{2}}=6000\frac{kg.m}{s^{2}}=6000N$

Hence, we have that the net force is equal to 6000N.

Have a nice day!

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Explain the following behaviour of molecules water rises up in a harrow tubes but mercury which is also a liquid falls in a narr

Mandarinka [93]

Answer: find the answer in the explanation

Explanation:

The capillarity of water molecules is different from the mercury molecules.

What is capillarity ?

This is the tendency of a liquid substance to rise in a capillary tube.

Molecules water rises up in a harrow tubes because of the force of adhesion between the water molecules and the tube molecules is greater than the force of cohesion between the water molecules. This helps water to wet the tube and rise. While mercury which is also a liquid falls in a narrow tubes to level below the outside surface because the force of cohesion between the mercury molecules is greater than the force of adhesion between the mercury molecules and the tube molecules. Mercury does not wet.

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4 years ago

You are most likely sitting in a room that is around 299 K. What is this temperature of the room in Celsius

Rom4ik [11]

Answer:

25.85 °C

Explanation:

To convert from ° C to K

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2 years ago

The displacement at any given time of an object is x = 6 sin 98t , where the symbols have their usual meanings. i). Proof that t

Fofino [41]

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation, which are represented in the given function as 6 units and 98 rad/s respectively.

<h3>General wave equation for simple harmonic motion</h3>

y = A sinωt

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A is amplitude of the motion
ω is angular frequency

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A = 6 units

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ω = 98 rad/s

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation. Thus, the wave is executing simple harmonic motion.

Learn more about simple harmonic motion here: brainly.com/question/17315536

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2 years ago

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77julia77 [94]

I’m pretty sure it’s C.

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3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$