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s2008m [1.1K]
4 years ago
9

express the following in metres (1) 52fm (2) 26 Mm (3)12am (4) 69 pm (5) 85 mm​

Physics
1 answer:
AlekseyPX4 years ago
7 0

Answer:........u are so sweet

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LVULAN
3241004551 [841]

For this case we have that by definition, the kinetic energy is given by the following formula:

k= \frac {1} {2} * m * v ^ 2

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

m = 100 \ kg\\v = 9 \frac {m} {s}

Substituting the values we have:

k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050

finally, the kinetic energy is 4050 \ J

Answer:

Option A

7 0
4 years ago
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o-na [289]

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3 0
3 years ago
Which of the following is an example of a medium for a visible light wave?
goldfiish [28.3K]

Answer:

C. green light

Explanation:

4 0
3 years ago
Read 2 more answers
Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

g=9.8 m/s^2 is the acceleration due to gravity

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

5 0
3 years ago
An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.
Rudik [331]
The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = </span>ρ\frac{l}{A}
where l = length of the wire
A = area of the wire
A = \pi r^{2} where, r = \frac{diameter of wire}{2}

Thus, on finding the ratio of resistance of the two wires, we get,

\frac{R1}{R2} =  \frac{l1A2}{l2A1}

here, R1 = R
l1 = 8m
l2 = 2m
A1=π0.25^{2}
A1=π0.50^{2}

we get. R2 = 16R
7 0
4 years ago
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