Why is doing research helpful if you're creating the procedure for an experiment?

Unit of energy density of electric field is a) NC-3 b) JC-3 c) Jm-3 d)JF

erik [133]

Answer:

the answer is c

8 0

4 years ago

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The value of gravitational acceleration of a body on earth is 9.8m/s^2. The gravitational potential energy for a 1.00 kilogram o

emmasim [6.3K]

Answer:

The object is 1.2755 meters above the ground

Explanation:

Recall the formula for gravitational potential energy for an object of mass "m" at a height "h" above the ground:

$PE=m\,*\,g\,*\,h$

In this case, since we are given the mass of the object and the object's potential energy, we can estimate the only unknown (height "h") from the formula shown above. Also since all units are given in the SI system, the result for the object's height will result in meters:

$PE=m\,*\,g\,*\,h\\12.5\,J=1.0\,kg\,*9.8\,\frac{m}{s^2} \,*\,h\\\frac{12.5}{9.8} \,meters= h\\h=1.2755\ meters$

4 0

3 years ago

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0

3 years ago

I KNOW YALL SEE THIS I NEED HELP GOD WILL GIVE U MANY BLESSING HELP A POOR SOUL OUT THE RECENT QUESTIONSSSSSSSSSSSSS. ://///

photoshop1234 [79]

Answer:

ok

Explanation:

will do

4 0

2 years ago

Read 2 more answers

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0

3 years ago