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Umnica [9.8K]
3 years ago
15

Measurements of acceleration are given in units of ?

Physics
2 answers:
kaheart [24]3 years ago
6 0
The measurement of acceleration are meters per a second
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Lady_Fox [76]3 years ago
3 0
Meters per second squared: \frac{m}{s^{2}}

If you think about it, acceleration is about how fast speed changes. Speed is measured in meters per second: \frac{m}{s}

So if you take that and just measure it over time, you get meters per second squared.
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A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant an
masya89 [10]

Answer:

Explanation:

Given

Wheels are rotating with constant angular velocity let say \omega

Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.

But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.

(a) yes, there will be tangential velocity which is given by

v=r\cdot \omega

where r=radial distance from center

(b)tangential acceleration

there would be no tangential acceleration as velocity is constant

(c)centripetal acceleration

Yes, there will be centripetal acceleration given by

a_c=\omega ^2\times r

                                   

7 0
3 years ago
Study the scenario.
Otrada [13]

If no other forces act on the object, according to Newton’s first law, the spacecraft will continue moving at a constant velocity, assuming that a planet or something with large mass doesn’t cross its path. Forces are not required to continue the motion of an object on a frictionless plane at a constant rate.

7 0
3 years ago
Read 2 more answers
What does the "coefficient of friction" tell you?
Arte-miy333 [17]
  • a coefficient of fraction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.
8 0
3 years ago
Question is attatched!
lina2011 [118]
All it does is lets him pull in a more convenient direction to raise the load. It has no effect on the required force.
4 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
4 years ago
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