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drek231 [11]
3 years ago
11

5. During the annual shuffleboard competition, Renee gives her puck an initial speed of 9.32 m/s. Once leaving her stick, the pu

ck slows down at a rate of -4.06 m/s/s. Determine the time it takes the puck to slow to a stop.
Physics
1 answer:
abruzzese [7]3 years ago
3 0
V=0 v²=0, A=v-u/t. T=v-u/a. T= 0-9.32/-4.06 therefore time = 2.296 seconds
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Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?​
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Answer:

i hope this will help you :)

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

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6 0
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A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric f
horrorfan [7]

Answer:

2271.16N/C  upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

W=mg\\g=9.81m/s^{2}

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

W=0.00382kg*9.81m/s^{2}

W=0.0375N

To calculate the electric field,

E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is  repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward

4 0
3 years ago
From this diagram which of the following can you conclude
Volgvan

Answer: A

Explanation:

I say opposites attract

3 0
3 years ago
Read 2 more answers
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Sveta_85 [38]

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8 0
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A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
3 years ago
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