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Scorpion4ik [409]
3 years ago
14

5) With a power of 0.12HP it is possible to move a piece of furniture from one place to another, in 14 seconds, applying a force

of 70N on it. Calculate: a- The work done. b- The distance traveled.
Physics
1 answer:
viva [34]3 years ago
7 0

Answer:

a) 1250 J

b) 17.9 m

Explanation:

Convert horsepower to watts:

0.12 HP = 89.5 W

a) Work = power × time

W = (89.5 W) (14 s)

W = 1250 J

b) Work = force × distance

1250 J = (70 N) d

d = 17.9 m

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Create a Graph : Include numbers, axes, and order pairs
dlinn [17]
Answer:

a) Peak value = 20

b) Average value = 9.0

c) RMS value = 14.15

Explanation:

The peak-to-peak value = 40

v_{p-p}=40\begin{gathered} v_{p-p}=2v_p \\  \\ 40=2v_p \\  \\ v_p=\frac{40}{2} \\  \\ v_p=20 \end{gathered}

c) The rms value

\begin{gathered} v_{rms}=\frac{v_p}{\sqrt{2}} \\  \\ v_{rms}=\frac{20}{\sqrt{2}} \\  \\ v_{rms}=14.14 \end{gathered}

b) Average value over alternation of the sine wave

\begin{gathered} v_{avg}=0.637v_p \\  \\ v_{avg}=0.637\times14.14 \\  \\ v_{avg}=9.0 \end{gathered}

6 0
1 year ago
A flywheel with radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.400 rad/s2.For a po
oksian1 [2.3K]

Answer:

0.12\ m/s^2

Explanation:

Given that,

The radius of a flywheel, r = 0.3 m

Angular acceleration of a flywheel, \alpha =0.4\ rad/s^2

We need to find the magnitude of the tangential acceleration after 2.00 s of acceleration.

The relation between the tangential and angular acceleration is given by :

a_t=r\alpha \\\\a_t=0.3\times 0.4\\\\a_t=0.12\ m/s^2

So, the required magnitude of tangential acceleration is 0.12\ m/s^2.

6 0
3 years ago
Two particles each have the same mass but particle #1 has four times the charge of particle #2. Particle #1 is accelerated from
marin [14]

Answer:

 v_2 = 2*v  

Explanation:

Given:

- Mass of both charges = m

- Charge 1 = Q_1

- Speed of particle 1 = v

- Charge 2 = 4*Q_1

- Potential difference p.d = 10 V

Find:

What speed does particle #2 attain?

Solution:

- The force on a charged particle in an electric field is given by:

                                       F = Q*V / r

Where, r is the distance from one end to another.

- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:

                                      F_net = m*a

- Equate the two expressions:

                                      a = Q*V / m*r

- The speed of the particle in an electric field is given by third kinetic equation of motion.

                                      v_f^2 - v_i^2 = 2*a*r

Where, v_f is the final velocity,

            v_i is the initial velocity = 0

                                      v_f^2 - 0 = 2*a*r

Substitute the expression for acceleration in equation of motion:

                                       v_f^2 = 2*(Q*V / m*r)*r

                                       v_f^2 = 2*Q*V / m

                                       v_f = sqrt (2*Q*V / m)

- The velocity of first particle is v:

                                       v = sqrt (20*Q / m)

- The velocity of second particle Q = 4Q

                                       v_2 = sqrt (20*4*Q / m)

                                       v_2 = 2*sqrt (20*Q / m)

                                       v_2 = 2*v  

3 0
3 years ago
What is a prediction
faust18 [17]
An educated guess about something. (What might happen in the future)
7 0
3 years ago
Read 2 more answers
7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf
Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to  1 μV/m

Intrinsic impedance of air = 120 π  Ω

Intrinsic impedance of  water = \dfrac{120\pi}{\epsilon_r}

Using equation to solve the problem

  E(z) = E_0 e^{-\alpha\ z}

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}

\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}

\alpha =21.07\ Np/m

now ,

  1 \times 10^{-6} = 20 e^{-21.07\times z}

  e^{21.07\times z}= 20\times 10^{6}

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0
3 years ago
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