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Strike441 [17]
3 years ago
7

30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

k in half as much time?
A) 12P
B) 6P
C) P
D) P
E) P
Physics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

P=\frac{W}{T}

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

P=\frac{W}{T}

Later, the force does six times more work, so the work done now is

W'=6W

And this work is done in half the time, so the new time is

T'=\frac{T}{2}

Substituting into the equation of the power, we find the new power produced:

P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P

So, 12 times more power.

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Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

                            =       zero            - (43 m/s)

                            =          -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

                                          =  (-43 / 0.28)  (m/sec) / sec

                                          =        153.57...  m/s²

                                          =        1.5...  x 10²  m/s²  .

5 0
3 years ago
Read 2 more answers
A 150 g air-track glider is attached to a spring. The glider is pushed in 8.80 cm and released. A student with a stopwatch finds
Alla [95]

Answer:

2.06 N/m

Explanation:

The system makes 10.0 complete oscillations in 17.0 s. So, the frequency of the system is

f=\frac{10.0}{17.0 s}=0.59 Hz

The angular frequency of the system is given by

\omega = 2\pi f=2\pi (0.59 Hz)=3.71 rad/s

In a simple harmonic motion, the angular frequency is related to the mass and the spring constant by

\omega=\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

Here we know

\omega=3.71 rad/s\\m = 150 g = 0.150 kg

So we can solve the formula to find k:

k=\omega^2 m = (3.71 rad/s)^2 (0.150 kg)=2.06 N/m

5 0
4 years ago
Choose all the answers that apply.
Gwar [14]

Answer: An independent variable is the variable which is changed. Experiments are usually published to communicate findings to the scientific community.

Disproving a hypothesis means it cannot be tested again.

Explanation: i hoped that helped.

4 0
3 years ago
A particle with negative charge q and mass m = 2.65×10−15 kg is traveling through a region containing a uniform magnetic field B
Norma-Jean [14]

Answer:

q = 2,95 10-6 C

Explanation:

The magnetic force on a particle is described by the equation

      F = q v x B

Where bold indicate vectors

Let's make the vector product

      vxB =\left[\begin{array}{ccc}i&j&k\\-3&4&12\\0&0&-0.13\end{array}\right]

                     

      v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]

      vx B = 1.20 106 [0.52 i ^ - 0.39j ^]

As they give us the force module, let's use Pythagoras' theorem,

     |v xB | =1.20 10⁶ √( 0.52² + 0.39²)

    |v x B| = 1.20 10⁶ 0.65

     v xB = 0.78 10⁶

Let's replace and calculate

    2.30 = q 0.78 10⁶

    q = 2.3 / 0.78 106

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4 0
3 years ago
An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11
Arada [10]

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

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