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DENIUS [597]
3 years ago
14

A glass falls off a shelf. How fast is the glass moving after 0.4 seconds?

Physics
2 answers:
Rufina [12.5K]3 years ago
8 0

Answer:

3.9 m/s

Explanation:

Apex ; good luck everyone! :)

tresset_1 [31]3 years ago
6 0

On Earth, gravity increases the speed of a falling object by about 9.8 m/s
every second that it falls.

In 0.4 second, its speed increases by (0.4 x 9.8) = 3.92 m/s.

If it just rolled off of the shelf at the beginning of that 0.4 second, then it
started from zero speed, and its speed after 0.4 second is 3.92 m/s.

Its velocity is 3.92 m/s downward.


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A particle in uniform circular motion requires a net force acting in what direction?
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The net force will point towards the acceleration of the object, as supported by Newton's second law.
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the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds
ki77a [65]

Answer:

Explanation:

The question is incomplete.

The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.

s= sin2(pi)t

Acceleration = d²S/dt²

dS/dt = 2πcos2πt

d²S/dt² = -4π²sin2πt

A(t) = -4π²sin2πt

Next is to find acceleration after 4.5 seconds

A(4.5) = -4π²sin2π(4.5)

A(4.5) = -4π²sin9π

A(4.5) = -4π²sin1620

A(4.5) = -4π²(0)

A(4.5) = 0m/s²

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3 years ago
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8 0
3 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

5 0
3 years ago
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