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3 years ago

1. A train increases its speed steadily from 10m/s to 20m/s in 1 minute.

Physics

2 answers:

Wewaii [24]3 years ago

5 0

The final speed is v

Setler [38]3 years ago

4 0

Explanation:

The final speed is

v=20m/s

The initial speed is

u=10m/s

The time is {eq}t = 1\,{\rm{min}} =...

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What's the difference between a wavelength and an an amplitude?

valkas [14]

Wavelength - the distance from one wave crest or trough to another wave crest or trough. Amplitude - the distance from the median point or "middle" of the wave straight up to a crest (a maximum) or straight down to a trough (or minimum), which is the peak amplitude; or the distance from a trough straight up to a crest, or a crest straight down to a trough, called peak-to-peak amplitude.

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4 years ago

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The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C, where U will be in joules w

satela [25.4K]

Answer:

F = - 2 A x - B

Explanation:

The force and potential energy are related by the expression

F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^

Where i ^, j ^, k ^ are the unit vectors on the x and z axis

The potential they give us is

U (x) = A x² + B x + C

Let's calculate the derivatives

dU / dx = A 2x + B + 0

The other derivatives are zero because the potential does not depend on these variables.

Let's calculate the strength

F = - 2 A x - B

3 0

3 years ago

Calculate the GPE of a rock with a mass of 55kg on a cliff that is 27m high

GaryK [48]

Answer:

P=mgh

P=mghm=55

P=mghm=55g=9.8 or ~10

P=mghm=55g=9.8 or ~10h=27

Explanation:

hope it will help you have a great day bye and Mark brainlist if the answer is correct

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 #carry on learning

4 0

2 years ago

How fast must a 1000 kg car be moving to have a kinetic energy of 2.0*10^3

lawyer [7]

Kinetic Energy is defined by Ke=1/2mv^2. Plug in and solve for v.

2,000 = 1/2(1000)(v)^2
4=(v)^2
v=2 m/s

The car must move at 2 m/s to have a Ke of 2,000 Joules.

6 0

3 years ago

A police officer in hot pursuit of a criminal drives her car through an unbanked circular (horizontal) turn of radius 300 m at a

Mamont248 [21]

Answer:

The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

Explanation:

Given:

Mass of the driver is, $m=55\ kg$

Radius of circular turn is, $R=300\ m$

Linear speed of the car is, $v=22.2\ m/s$

Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.

The forces are:

1. Weight = $mg=55\times 9.8=539\ N$

2. Centripetal force, 'F', which is given as:

$F=\frac{mv^2}{R}\\F=\frac{55\times (22.2)^2}{300}\\\\F=\frac{55\times 492.84}{300}\\\\F=\frac{27106.2}{300}=90.354\ N$

Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,

$\tan \theta=\frac{90.354}{539}\\\tan \theta=0.1676\\\theta=\tan^{-1}(0.1676)=9.52\approx 10$ °

Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

3 0

3 years ago