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2 years ago

The ball rolling along a straight and level path. The ball is rolling at a constant

Physics

1 answer:

Aleksandr-060686 [28]2 years ago

6 0

Answer:6 joules

Explanation:

Mass(m)=3kg

Velocity(v)=2m/s

Kinetic energy=0.5 x m x v^2

Kinetic energy=0.5 x 3 x 2^2

Kinetic energy=0.5 x 3 x 2 x 2

Kinetic energy=6

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A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

2 years ago

Which shows a decrease in fluid pressure? A. A fan is turned from high speed to low speed. B. Oxygen is compressed as it is put

Volgvan

Answer:

Option A.

A fan is turned from high speed to low speed.

Explanation:

It is important to note that air is also a fluid.

In a system, static pressure of air increases with the speed of rotation of the fan. This is because when the speed of the fan is increased, the force with which it is pushing the air molecules is increased. Since pressure is a relationship between force and area, the pressure of the air molecules will be increased.

Conversely, when the speed of the fan is reduced, the priming force on the air molecules will be reduced, hence the pressure of the air will drop.

This makes option A the correct option

8 0

2 years ago

Read 2 more answers

There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot fl

nikdorinn [45]

Answer:

Also 3s.

Explanation:

Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.

8 0

2 years ago

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000 cycles

7 0

2 years ago

An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init

elena-14-01-66 [18.8K]

Answer:

Approximately $71\%$ .

Explanation:

The formula for the kinetic energy $\rm KE$ of an object is:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ ,

where

$m$ is the mass of that object, and
$v$ is the speed of that object.

Important: Joule ( $\rm J$ ) is the standard unit for energy. The formula for $\rm KE$ requires two inputs: mass and speed. The standard unit of mass is $\rm kg$ while the standard unit for speed is $\rm m \cdot s^{-1}$ . If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

$m = 63\; \rm g = 0.063\; \rm kg$ .

Initial $\rm KE$ of this arrow:

$\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}$ .

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

$\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%$ .

8 0

2 years ago