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3 years ago

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A 4 kg kitten is sliding at 12 m/s on a horizontal frictionless surface. A constant force is applied that slows it with an accel

eration of 3 m/s/s. How much work must this force do to stop the kitten

1 answer:

Liono4ka [1.6K]3 years ago

3 0

Answer:

288 Joules

Explanation:

Work= delta kinetic energy. To find the work required to stop the kitten, you have to find the delta kinetic energy between when the kitten is moving and when it is stopped. Since it is not moving when it's stopped, it has no kinetic energy. So to find the work, just find the kinetic energy of the kitten moving at 12m/s.

KE=1/2((m)(v^2)), therefore

delta KE=(1/2)*((4kg)*((12m/s)^2))-0

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You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double

dimulka [17.4K]

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

$\tan\theta=\dfrac{y}{D}$

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

$\tan\theta=\dfrac{1.52}{120}$

$\theta=\tan^{-1}\dfrac{1.52}{120}$

$\theta=0.725$

We need to calculate the wavelength

Using formula of wavelength

$d\sin\theta=n\lambda$

Put the value into the formula

$0.025\times\sin0.725=5\times\lambda$

$\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}$

$\lambda=6.326\times10^{-7}\ m$

$\lambda=633\ nm$

Hence, The wavelength of the light is 633 nm.

4 0

3 years ago

(1) Which appliance is designed to transfer electrical energy to kinetic energy?

sweet-ann [11.9K]

Answer:

bb kettle

Explanation:

it transfres electricsl to kinetic

8 0

3 years ago

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p

aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$ E

and, we know that, F = ma

So,

$m_{p}$ a = $q_{p}$ E

a = $\frac{q_{p}.E }{m_{p} }$ Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2 $at^{2}$

Here, u = 0.

S = 1/2 $at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ( $\frac{q_{p}.E }{m_{p} }$ ) $t^{2}$ Equation 2

So,

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ( $\frac{q_{e}.E }{m_{e} }$ ) $t^{2}$ Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S $m_{p}$ = $m_{e}$ D - $m_{e}$ S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm

7 0

3 years ago

A heat engine exhausts 3 000 J of heat while performing 1 500 J of useful work. What is the efficiency of the engine

amid [387]

efficiency=work output/work input×100

since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input

efficiency=1500/3000×100

=50%

3 0

3 years ago

The multi-link mechanism in the Variable Compression Turbo Engine _____, thus varying the compression ratio.

Fittoniya [83]

Answer:

A. Adjusts how far down the piston travels

Explanation:

This type of engine changes the possition of the piston in order to modify the compression chamber volume and therefore the compression ratio of the engine. The volume of the chamber is proportional to the run of the piston (how far down the piston travels)

This engine is used to achive the optimal compression rate in each individual stage.

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3 years ago