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Ray Of Light [21]
3 years ago
9

Please help.......

Physics
1 answer:
IceJOKER [234]3 years ago
3 0

Answer:

(L: Length, T: Time)

p: Dimension: L; unit: m

q: Dimension: L/T or (L)*(T)^-1; unit: m/s

r: Dimension: L/T^2 or (L)*(T)^-2; unit: m/s^2

Explanation:

since y is distance (Length), make all terms L distance.

p is same as y dimension ==> dimension: L; unit: m (meter)

qt dimension is L ==>  q dimension :L/T; unit: m/s

rt^2 dimension is L ==> r dimension : L/T^2; unit: m/s^2

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A car is travelling with a velocity of 10 m/s and has a mass of 550 kg. The<br> car has<br> energy.
Elenna [48]

Answer: The car has a kinetic energy (because it's in motion) of:  27500J

Explanation:

Formula: E=\frac{1}{2}mv^2

E=\frac{1}{2}mv^2

E=\frac{1}{2}(550kg)(10m/s)^2

E=\frac{1}{2}(550kg)(100m^2/s^2)

E=\frac{1}{2}(55000J)\\E=27500J

3 0
3 years ago
If we compare the force of gravity too strong nuclear force we would conclude that
Yuri [45]
The force of gravity is much weaker than the strong nuclear force. But the strong nuclear force only acts over short distances, such as within the nuclues. The gravitational force can act over infinite distance.

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3 years ago
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3 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
3 years ago
You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

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b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
4 years ago
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