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3 years ago

A block sliding across a level surface has a mass of 2.5 kg and a mechanical energy of 20 joules. What is its velocity?

1 answer:

8 0

Hi!

The energy of the block is 4 m/s

To calculate this, you need to use the equation for kinetic energy. The block is sliding (i.e. it's moving). If the object is sliding across a level surface, the only energy it has is kinetic energy, because there is no change in potential energy (which changes with height). So, the mechanical energy will be pure kinetic energy. The equation is the following, derived from the expression for kinetic energy:

$v= \sqrt{ \frac{2*Ke}{m}}=\sqrt{ \frac{2*20 (kg*m^{2}*s^{-2}) }{2,5kg}}=4 m/s$

Have a nice day!

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How are the fiducial points of the Celsius and Fahrenheit scales similar?

taurus [48]

The fiducial points of the Celsius and the Fahrenheit temperature scales are the boiling and freezing points of pure water at 1 atm of pressure.

In short, Your Answer would be Option D

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7 0

3 years ago

A person measures his or her heart rate by counting the number of beats in 30s. If 40±1 beats are counted in 30.0±0.5s, what is

nordsb [41]

Answer:

$Rate = 1.33 \pm 0.055$ beats per second

Explanation:

Number of heart beats = $40 \pm 1$

time taken = $30.0 \pm 0.5 s$

now we have

$N = 40 \pm 2.5$ %

$t = 30.0 \pm 1.67$ %

now rate of heart beat is defined as number of heart beat per unit of time

so we have

$Rate = \frac{N}{t}$

$Rate = \frac{40 \pm 2.5}{30 \pm 1.67}$

so we have

$Rate = 1.33 \pm (2.5 + 1.67 )$

$Rate = 1.33 \pm 4.17$ %

$Rate = 1.33 \pm 0.055$ beats per second

7 0

3 years ago

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3 years ago

A cyclist going downhill is accelerating at 1.2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is t

Blababa [14]

Answer:

Initial Velocity is 4 m/s

Explanation:

What is acceleration?

It is the change in velocity with respect to time, or the rate of change of velocity.

We can write this as:

$a=\frac{\Delta v}{t}$

Where

a is the acceleration

v is velocity

t is time

$\Delta$ is "change in"

For this problem , we are given

a = 1.2

t = 10

Putting into formula, we get:

$a=\frac{\Delta v}{t}\\1.2=\frac{\Delta v}{10}\\\Delta v = 1.2*10\\\Delta v = 12$

So, the change in velocity is 12 m/s

The change in velocity can also be written as:

$\Delta v = Final \ Velocity - Initial \ Velocity$

It is given Final Velocity = 16, so we put it into formula and find Initial Velocity. Shown Below:

$\Delta v = Final \ Velocity - Initial \ Velocity\\12=16-Initial \ Velocity\\Initial \ Velocity = 16 - 12 = 4$

hence,

Initial Velocity is 4 m/s

3 0

4 years ago

Read 2 more answers

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago