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Anarel [89]
3 years ago
8

________ is a force acting through distance.

Physics
1 answer:
Andreas93 [3]3 years ago
8 0
High school???
No way
It's work.
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In an arcade game a 0.099 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas
Sonja [21]

Answer:

The speed of disk is 1.98 \frac{m}{s}

Explanation:

Given:

Mass of m = 0.099 kg

Spring constant k = 244 \frac{N}{m}

Compression of spring x = 4 \times 10^{-2} m

From energy conservation theorem,

Spring potential energy converted into kinetic energy,

   \frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}

  v = \sqrt{\frac{k x^{2} }{m} }

  v = \sqrt{\frac{244 \times 16 \times 10^{-4} }{0.099} }

  v = 1.98 \frac{m}{s}

Therefore, the speed of disk is 1.98 \frac{m}{s}

8 0
3 years ago
"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after
wel

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

3 0
3 years ago
If an automobile engine delivers 42.0 hp of power, how much time will it take for the engine to do 6.20 â 105 j of work? (hint:
Elena L [17]
To be able to answer this item, we are to calculate the power that the machine could deliver from hp to kW. 

      (45 hp)(746 W/1 hp) = 33570 W

Power is the amount of energy delivered at a certain period. 

             t = (6.20 x 10^2 J)/ (33570 kJ/s)

             t = 0.01845 s
7 0
3 years ago
a sensor light installed on the edge of a home can detect motion for a distance of 50 feet in front and with a range of motion o
Dvinal [7]

Answer:

4363.3231 feets²

Explanation:

Given that :

Distance, r = 50 ft

θ = 200°

The arc length of area covered :

Arc length = θ/360° * πr²

Arc length = (200/360) * 50 ft ^2 * π

Arc length = 0.5555555 * 2500 * π

Arc length = 4363.3231 feets²

7 0
2 years ago
15) What is the frequency of a pendulum that is moving at 30 m/s with a wavelength of .35 m?
____ [38]

A pendulum is not a wave.

-- A pendulum doesn't have a 'wavelength'.

-- There's no way to define how many of its "waves" pass a point
every second.

--  Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.

-- The frequency of a pendulum depends only on the length
of the string from which it hangs.


If you take the given information and try to apply wave motion to it:

             Wave speed = (wavelength) x (frequency)

             Frequency  =  (speed) / (wavelength) ,

you would end up with

             Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz

Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ?     That's pretty absurd. 

This math is not applicable to the pendulum.

6 0
2 years ago
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