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amid [387]
3 years ago
8

Is the net force on an object moving with uniform velocity zero or not zero?

Physics
1 answer:
NikAS [45]3 years ago
3 0

Answer:

Zero

Explanation:

The net force acting on an object moving with uniform velocity is zero. This is an accordance with the Newton's first law of motion.

Newton's first law of motion states that a body will continue in its state of rest or uniform motion unless it is acted upon by an external force.

Since this body is moving with uniform motion, it is not accelerating. When a body is acceleration, the net force on it is not zero.

But this one is moving with uniform motion. The net force on the body is balanced and zero.

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An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
An object with a non-zero speed must be _________.
pogonyaev
It would be either A or C if its still moving and not stopping
8 0
3 years ago
Read 2 more answers
A frog jumps for 4.0 seconds at a maximum horizontal distance of 0.8m. what is its velocity along the road?
Lerok [7]

Answer:

The frog's horizontal velocity is 0.2 m/s.

Explanation:

To solve this problem, we must first remember what velocity is and how we solve for it.  Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance.  If we plug in the given numbers for these variables and solve, we get the following:

v = x/t

v = 0.8m/4s

v = 0.2 m/s

Therefore, the correct answer is 0.2 m/s.  We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.

Hope this helps!

3 0
2 years ago
What is the path of a projectile called? Friction Track Trajectory acceleration
sergiy2304 [10]
It is trajectory acceleration. A friction track is a device to study motion in low friction environments, I believe. Does this help?
3 0
3 years ago
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A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected
Ganezh [65]

Answer:

The ball's initial kinetic energy

The ball comes to a stop at B. At this point its initial kinetic energy is converted into potential energy

Explanation:

A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.

From conservation of energy which states that energy can neither be created nor be destroyed, but can be transformed from one form to another.

Ki+Ui=Kf+Uf

Ki=initial kinetic energy

Ui=initial potential energy

Kf=final kinetic energy

Uf=final potential energy

we know that \frac{1}{2} mu^{2} +mgha=\frac{1}{2} mv^{2} +mghb

m=mass of the ball

ha=downward height a

hb=upward height b

u=initial velocity u

v=final velocity v, which is 0

g=acceleration due to gravity

v=0 at final velocity

1/2mu^2+mgha=0+1/2mv^2

ha=hb+Ki/mh

From the above equation, we can conclude that the ball's initial kinetic energy  is responsible for making the ball reach point B.

Point B is higher than point A from the motion gained by the ball

3 0
3 years ago
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