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3 years ago

What’s the unit for mass

Physics

1 answer:

KATRIN_1 [288]3 years ago

8 0

the unit for mass is kilogram

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Explanation:

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3 years ago

Which air mass would produce cold dry weather in the winter?

Solnce55 [7]

It is Continental polar ( only if this is for apex)

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3 years ago

Read 2 more answers

Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they h

omeli [17]

Answer:

Explanation:

Considering non - relativistic approach : ----

Speed of electron = 1 % of speed of light

= .01 x 3 x 10⁸ m /s

= 3 x 10⁶ m /s

Kinetic energy of electron = 1/2 m v²

= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²

= 40.95 x 10⁻¹⁹ J

Kinetic energy in electron comes from lose of electrical energy equal to

Ve where V is potential difference under which electron is accelerated and e is electronic charge .

V x e = kinetic energy of electron

V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹

V = 25.6 Volt .

6 0

3 years ago

A thin walled spherical shell is rolling on a surface. What

ExtremeBDS [4]

Answer:

$=\frac{1/3}{5/6} = 0.4$

Explanation:

Moment of inertia of given shell $= \frac{2}{3} MR^2$

where

M represent sphere mass

R -sphere radius

we know linear speed is given as $v = r\omega$

translational $K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2$

rotational $K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

total kinetic energy will be

$K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2$

$K.E =\frac{5}{6} MR^2 \omega^2$

fraction of rotaional to total K.E

$=\frac{1/3}{5/6} = 0.4$

8 0

3 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

3 years ago