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3 years ago

Which of the following is not a an example of dissipated energy?

1 answer:

4 0

Which of the following is not a an example of dissipated energy?
b. kinetic

When energy is changed from one form to another, ____.

b. all of the energy can be accounted for

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What is mesothermal climates

Snowcat [4.5K]

Answer:

Mesothermal regions have moderate climates. They are not cold enough to sustain a layer of winter snow but are also not remain warm enough to support flowering plants (and, thus, evapotranspiration) all year.

Explanation:

Please give the brainliest.

3 0

1 year ago

A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

scZoUnD [109]

A charged particle moving in a magnetic field experiences a force equal to:

$\vec{F}=q\vec{v}\times \vec{B}$

Thus, the magnitude of the force that the proton experiences is given by:

$F=qvBsin\theta$

The magnetic field is perpendicular to the proton's velocity, therefore, we have $\theta=90^\circ$ . Replacing the given values, we obtain:

$F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N$

3 0

3 years ago

A bird flies from the South Pole to the North Pole. Part of the journey is 1000 miles that takes 2 weeks. What is the bird’s vel

viktelen [127]

1000 miles = 1610km = 1.61x10^6m
2 weeks = 14 days = 14x24x1440

V=d/t = 1.61x10^6/14x24x1440
= 3.33m/s

6 0

3 years ago

Read 2 more answers

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

7 0

3 years ago

A block of mass 2 kg slides down a frictionless ramp of length 1.3 m tilted at an angle 25o to the horizontal. At the bottom of

marin [14]

Answer:

Diagrams in pictures

Explanation:

Using energy I can get

m g h = 1/2 m v^2

So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.

(H= 1,3m sin25)

V=2,32m/s

V= a t

And

X= v t +1/2 a t^2

Knowing v=2,32 m/s and x= 1,3 m

I can get

a= 6,21m/s2

F= m a

I can get the force of the box when it collides with the spring

F= 12, 425 N

The force the spring makes on the box then is

F = -12,425N = -k d

Then the spring's constant is k= 51,75N/m

To make the two diagrams I need the functions of time when the box slows down

I use the same two equations

V= a t

And

X= v t + 1/2 a t^2

Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.

I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).

Then,

V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec

X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.

for time between 0 and 0,689 sec

Diagrams and equations are in the pictures

7 0

3 years ago