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seraphim [82]
3 years ago
11

What is a real world application of atwood's machine?

Physics
1 answer:
Darina [25.2K]3 years ago
8 0

The real world application of the atwood’s machine is to be able to demonstrate specific principles such as the acceleration and dynamics in a way to understand and know how it is being demonstrated in the real world and in different situations.

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The most common units for expressing the density of a substance are the g/cm3.
Luden [163]
True.

Density = mass / volume,  Unit = g / cm³.

This is a common unit because of its affiliation with the SI unit and because that also our popular liquid which is water = 1 g/cm³  
6 0
3 years ago
A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.
Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

P = \frac{W}{t}

W= F*s

Therefore,

P=\frac{F*s}{t}

Substitute the given values of force , displacement and time

F = 124 N,s = 32.0 m,t = 8.0 s

P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W

Thus the Power can be rounded off to the nearest value of 500 W

3 0
3 years ago
Read 2 more answers
What is the speed of a truck that travels 10 i’m in 10 minutes ?
fiasKO [112]

10km/10min is a legitimate speed. So is meters/sec, km/hour (kph), etc.  

Kph is very common for vehicles:  

10 km/10 min (60 min/hr) = 60 kph

7 0
4 years ago
Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro
luda_lava [24]

Answer:

c. 4.582\times10^{21} kg

Explanation:

r_{i} = Initial distance between asteroid and rock = 7514 km = 7514000 m

r_{f} = Final distance between asteroid and rock = 2823 km = 2823000 m

v_{i} = Initial speed of rock = 136 ms⁻¹

v_{f} = Final speed of rock = 392 ms⁻¹

m = mass of the rock

M = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg

8 0
4 years ago
Sound waves, water waves, and light waves are all alike in that they all
Paha777 [63]
Sound and water waves are longitudinal waves, they require a medium to travel through and occilate particles 90 degrees to the wave motion
Light is a transverse wave. It doesnt require a medium to travel through.
All three reflect, refract and diffract


Light is difficult to think of because it acts in ways which waves cannot explain in some cirumstances. It acts like a particle (called photons) in some conditions, but acts like a normal sound or water wave does in others. Try not to get too caught up in light being a wave or a particle because even physists dont know how to explain it yet.
7 0
3 years ago
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