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3 years ago

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Physics

1 answer:

GenaCL600 [577]3 years ago

7 0

Answer:

I don't know about these problems

Explanation:

I don't know about these problems at all.

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a lawn mower is pushed with a force of 50n. if the angle between the handle of the mower and the ground is 30°. why doesn't the

Mademuasel [1]

Answer:

IT doesnt push down because the lawnmower has wheels and the ground is hard

Explanation:

Branliest

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3 years ago

You're driving in a car at 50 km/h and bump into a car ahead traveling at 48 km/h in the same direction. the speed of impact is

salantis [7]

To solve this problem, we must remember about the law of conservation of momentum. The initial momentum mist be equal to the final momentum, that is:

m1 v1 + m2 v2 = (m1 + m2) v’

where v’ is the speed of impact

Since we are not given the masses of each car m1 and m2, so let us assume that they are equal, such that:

m1 = m2 = m

Which makes the equation:

m v1 + m v2 = (2 m) v’

Cancelling m and substituting the v values:

50 + 48 = 2 v’

2 v’ = 98

v ‘ = 49 km/h

<span>The speed of impact is 49 km/h.</span>

6 0

3 years ago

A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee

SOVA2 [1]

Answer:

$\omega=0.12\frac{rad}{s}$

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

$\omega=\frac{2\pi}{T}(1)$

Here T is the period, that is, the time taken to complete onee revolution:

$T=\frac{2\pi r}{v}(2)$

Replacing (2) in (1):

$\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}$

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4 years ago

Alonzo sprints for 500 meters along a straight track during a race. After crossing the finish line, he to walks back along the t

ehidna [41]

450 meters is the displacement

8 0

4 years ago

Read 2 more answers

Calculate the number of free electrons and holes (in m-3) in an intrinsic semiconductor that has electron and hole mobilities of

allochka39001 [22]

Answer:

ni = 2.04e19

Explanation:

we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have

n = p = ni

from intrinsic carrier concentration

$\sigma = n\left | e \right | \mu_e + n\left | e \right | \mu_h$

$\sigma = ni\left | e \right | \mu_e + ni\left | e \right | \mu_h$

$\sigma = ni \left | e \right | ( \mu_e + \mu_h)$

1.7 = ni * 1.6*10^{-19} * (.35 + .17)

ni = 2.014 *10^{19} m^{-3}

ni = 2.04e19

5 0

3 years ago