Ask question

Ask question

All categories

3 years ago

5

A 6.0-kilogram block slides along a horizontal surface. If μk = 0.20 for the block and surface, at what rate is the friction for

ce doing work on the block at an instant when its speed is 4.0 m/s?

1 answer:

kupik [55]3 years ago

6 0

Answer:

Power = 47.0 Watt

Explanation:

As we know that friction force is given by

$F_f = \mu mg$

now we have

$\mu = 0.20$

m = 6.0 kg

now we have

$F_f = 0.20(6.0)(9.80) = 11.76 N$

now since we need to find the rate of work done by friction force

so we can say rate of work done is power due to friction force

so it is given as

$P = F_f (v)$

$P = 11.76 (4.0)$

$P = 47.0 Watt$

You might be interested in

The SI unit for energy is Newton true or false

ipn [44]

Explanation:

false....it is joules

..........

6 0

3 years ago

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of 1.0 nC/mm2 . If the potential between t

Archy [21]

Answer:

$A=81mm^2$

Explanation:

We know that for a capacitor $Q=CV$ , where <em>Q </em>is the charge of one plate, <em>C</em> the capacitance and <em>V</em> the potential between the plates.

We also know that $Q=\sigma A$ , since $\sigma$ is the surface charge density and <em>A</em> the area of the plate (both equal in our case).

Putting all together:

$A=\frac{CV}{\sigma}$

Which for our values is:

$A=\frac{(3\times10^{-12}F)(27\times10^3V)}{1\times10^{-9}C/mm^2}=81mm^2$

Where we notice that the S.I. units combination FV/C must not have units (we can verify it directly from their definitions or we notice that $mm^2$ is enough to describe an area).

8 0

3 years ago

Question 18/55 (2 p.)

Readme [11.4K]

Answer:

the correct answer is C v = 60 cm / s

Explanation:

The speed of a wave is related to the frequency and the wavelength

v = λ f

They indicate that the object performs 20 oscillations every second, this is the frequency

f = 20 Hz

the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength

λ = 3 cm = 0.03 m

let's calculate

v = 20 0.03

v = 0.6 m / s

v = 60 cm / s

the correct answer is C

6 0

3 years ago

A 2.35-kg rock is released from rest at a height of 21.4 m. Ignore air resistance and determine (a) the kinetic energy at 21.4 m

kvasek [131]

Explanation:

Given that,

The mass of rock, m = 2.35-kg

It was released from rest at a height of 21.4 m.

(a) The kinetic energy is given by : $E_k=\dfrac{1}{2}mv^2$

As the rock was at rest initially, it means, its kinetic energy is equal to 0.

(b) The gravitational potential energy is given by : $E_p=mgh$

It can be calculated as :

$E_p=2.35\times 9.8\times 21.4\\\\E_p=492.84\ J$

(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,

M = 0 J + 492.84 J

M = 492.84 J

Hence, this is the required solution.

6 0

4 years ago

A child and sled with a combined mass of 58.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed o

erma4kov [3.2K]

Answer:

The height is $h = 0.5224 \ m$

Explanation:

From the question we are told that

The combined mass of the child and the sled is $m = 58.0 \ kg$

The speed of the sled is $u = 3.20 \ m/s$

Generally applying SOHCAHTOA on the slope which the combined mass is down from

Here the length of the slope(L) where the combined mass slides through is the hypotenuses

while the height(h) of the height of the slope is the opposite

Hence from SOHCAHTOA

$sin (\theta) = \frac{h}{L}$

=> $Lsin(\theta) = h$

Generally from the kinematic equation we have that

$v^2 = u^2 + 2aL$

Here the u is the initial velocity of the combined mass which is zero since it started from rest

and a is the acceleration of the combined mass which is mathematically evaluated as

$a = g * sin (\theta )$

$v^2 = u^2 + 2 * g * sin (\theta )L$

=> $2Lsin(\theta ) = \frac{v^2 - u^2 }{g}$

=> $h = \frac{ v^2 - u^2}{2g}$

=> $h = \frac{ 3.20^2 - 0^2}{2 * 9.8 }$

=> $h = 0.5224 \ m$

6 0

3 years ago