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kari74 [83]
3 years ago
10

What determines the ideal mechanical advantage of a pulley?

Physics
1 answer:
leva [86]3 years ago
6 0
The ideal mechanical advantage of a compound pulley is
the number of strands of rope supporting the load.
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1.)State which definition of power below is correct.
Natalija [7]

C

kW, watts, kilowatts, W

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3 years ago
An elephant produces a 10Hz sound wave. Assuming the speed of sound in air is 344/s, determine the wavelength of this infrasonic
user100 [1]

Answer:

34.5 m

Explanation:

Given data : An elephant produces 10 Hz sound wave. Assuming the speed of sound in air is 345 m/s. To find : What is the wavelength of sound ? Answer : The wavelength of the sound is 34.5 m.

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2 years ago
An object appears red when viewed with white light. What color would it appear when viewed with blue light?
zvonat [6]

Answer:

Colour of any object is by the colour of light it reflects i.e if white light is incident on the object,it will reflect blue color.so it will appear blue.But if red light is incident on it,it will not reflect that and absorb it.so as it will not reflect any light it will appear black.

Explanation:

7 0
3 years ago
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Estimate how long the sun would last if it were merely a huge fire that was releasing chemical energy. Assume that the sun begin
stepan [7]

Answer:

≅ 17000 years or 1.7 x 10⁴ years

Explanation:

time= total energy/power

=  (10⁸J/kg)(2x10³⁰ kg) / 3.8 x 10²⁶ J/s

 = 526,315,789,473 s

=  16689 years

≅ 17000 years or 1.7 x 10⁴ years

7 0
2 years ago
A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a
scZoUnD [109]

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
T_h = \frac{T}{2}

Now, we can determine the velocity at which the can was launched at using the following equation:
v_f = v_i + at

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}

***vsinθ is the vertical component of the velocity.

Solve for 'v':
vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}

Now, recall that:
W = \Delta KE = \frac{1}{2}m(\Delta v)^2

Plug in the expression for velocity:
W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}

B.

We can use the same process as above, where T' = 2T and Th = T.

v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}

C.

The work done in part B is 4 times greater than the work done in part A.

\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}

4 0
2 years ago
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