What is the english system of measurement called?

VladimirAG [237]

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)

$\Delta t=8 s$

Then:

$a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}$

$a_{ave}=0.5 m/s^{2}$

5 0

3 years ago

What is the acceleration of a 10 kg mass pushed by a 5 N force?

yanalaym [24]

$g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}= \\ g=0,5 \frac{N}{kg}$

8 0

4 years ago

Read 2 more answers

Rays of light strike a bumpy road and are reflected.

ser-zykov [4K]

Answer:

last option is the correct one

6 0

3 years ago

Read 2 more answers

Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers

Dmitrij [34]

Answer:

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

6 0

4 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

4 years ago