All categories

4 years ago

What does the path of an object look like when it is in uniform motion ?

Physics

1 answer:

alexandr1967 [171]4 years ago

5 0

In uniform motion, the path is a straight line, and the object r moving along it at a constant speed.

You might be interested in

An engine extracts 452.8kJ of heat from the burning of fuel each cycle, but rejects 266.7 kJ of heat (exhaust, friction,etc) dur

Svetlanka [38]

Answer: The thermal efficiency of the engine is 41.09 %.

Explanation:

Efficiency is the ratio of the useful work performed to the total energy expended or heat taken in.

Formula for thermal efficiency of engine is

$\eta=1- \frac{Q_2}{Q_1}\times 100$

$\eta$ = efficiency

${Q_2}$ = heat rejected = 266.7 kJ

${Q_1}$ = heat extracted = 452.8 kJ

Putting in the values we get:

$\eta=1- \frac{266.7 kJ}{452.8 kJ }\times 100$

$\eta=0.41\times 100$

$\eta =41.09\%$

The thermal efficiency of the engine is 41.09 %.

7 0

4 years ago

Which of the following properties of materials does NOT affect resistance?

Shalnov [3]

Answer:

B thickness

Explanation:

thickness

4 0

3 years ago

swings a 5.5 kg cup of water in a vertical circle of radius 1.9 m. (a) What minimum speed must the cup have in this demo if the

Tanzania [10]

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

6 0

3 years ago

A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 270 kg · m2 and is rotating at 8.0 rev/min about a

Arte-miy333 [17]

Answer:

$\dot n = 6.042\,rpm$

Explanation:

The final angle speed of the merry-go-round is determined with the help of the Principle of Angular Momentum Conservation:

$(270\,kg\cdot m^{2})\cdot \left(8\,rpm\right) = [270\,kg\cdot m^{2}+(27\,kg)\cdot (1.80\,m)^{2}]\cdot \dot n$

$\dot n = 6.042\,rpm$

3 0

3 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.6 cm and diameter 2.95 cm fr

slavikrds [6]

Answer:

48.4293354946 N

Yes

Explanation:

d = Diameter of rod = 2.95 cm

h = Length of rod = 92.6 cm

$\rho$ = Density of rod = 7800 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Volume of rod

$V=\dfrac{1}{4}\pi d^2h\\\Rightarrow V=\dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}$

Mass is given by

$m=\rho V\\\Rightarrow m=7800\times \dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}\\\Rightarrow m=4.93673144695\ kg$

Weight is given by

$W=mg\\\Rightarrow W=4.93673144695\times 9.81\\\Rightarrow W=48.4293354946\ N$

The weight of the rod is 48.4293354946 N

The mass of the rod is 4.93673144695 kg which is light. So, I will be able to carry the rod without a cart.

7 0

3 years ago