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4 years ago

At what speed must a 950-kg subcompact car be moving to have the same kinetic energy as a 3.2×104−kg truck going 20 km/h?

Physics

1 answer:

Ostrovityanka [42]4 years ago

6 0

Answer:

950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

Explanation:

Kinetic energy of body with mass m and moving at velocity v is given by

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

KE of truck $= \frac{1}{2}*3.2*10^4*20^2$

KE of subcompact car $= \frac{1}{2}*950*v^2$

Equating

$\frac{1}{2}*3.2*10^4*20^2= \frac{1}{2}*950*v^2\\ \\ v=116.08 km/hr$

So 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

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What is weight in Newton’s, of a 50.-kg person on earth

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Answer:

It is about 490 Newtons. 490.3325 to be exact

Explanation:

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3 years ago

An EM wave has a wave length of 956 m, What type of electromagnetic waves is this wave

Assoli18 [71]

The electromagnetic spectrum allows finding the result for the question of type of electromagnetic wave has a wavelength of 956 nm

The wave is in the near infrared range.

The electromagnetic wave is an oscillation of electric fields and magnets field that sustain each other, this wave time has the characteristics that it can travel without a material medium, for which they are of extreme importance.

Consequently the spectrum of electromagnetic waves to each part has been given a different name second to its use, for example

Range Name

Ultraviolet light 200 to 400 nm

Visible light 400 to 700 nm

Infrared light (heat ) 700 nm to 1 mm

The latter is subdivided into:

Near infrared 700 nm to 1000 nm (1 μm)

Medium infrared 1 μm to 30 μm

Far infrared 50 μm to 1000 μm

It indicates that the wavelength of the measured radiation is 956 nm, therefore it is in the near infrared range.

Learn more about the electromagnetism spectrum here: brainly.com/question/23727978

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2 years ago

Atoms contain empty space true or false

olga nikolaevna [1]

The answer would be false

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3 years ago

Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

$f=\frac{qB}{2\pi m}$

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

$m=9.31\cdot 10^{-31} kg$ is the mass of the electron

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

The flux is calculated as

$\Phi = BA sin \theta$

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

$\theta=18^{\circ}$ is the angle between the direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

$A=(0.15 m)(0.08 m)=0.012 m^2$

So the flux is

$\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb$

See the last 7 answers in the attached document.

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5 0

3 years ago

A speedboat moves on a lake with initial velocity vector v1,x = 8.57 m/s and v1,y = -2.61 m/s, then accelerates for 6.67 seconds

KengaRu [80]

First, you find the velocity at each component. The general equation is:

a = (v2 - v1)/t

a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s

a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s

To find the final speed, find the resultant velocity by taking the hypotenuse.

v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s

3 0

3 years ago