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Nonamiya [84]
3 years ago
14

Find the kinetic energy of a 0.1 kg toy truck moving at a speed of 1.1 m/s

Physics
1 answer:
enot [183]3 years ago
8 0
0.0605J is your answer. Use the formula KE=1/2mv^2
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What are similarities and differences of the 1st, 2nd, and 3rd class levers?
lbvjy [14]
Similarities:
-- All three classes have a fulcrum (pivot).
-- All three classes have a point where the effort force is applied.
-- All three classes have a point where the load or resistance force is applied.
-- If you can find a place to stand and a lever that's long enough,
then you can move the Earth with a 1st or 2nd Class lever.

Differences:
-- The mechanical advantage of a 1st Class lever
can be greater than 1, equal to 1, or less than 1.
-- The mechanical advantage of a 2nd Class lever is always more than 1 .
-- The mechanical advantage of a 3rd Class lever is always less than 1 .



7 0
3 years ago
A 75 kg refrigerator is located on the 70th floor of skyscraper (300 meters above ground). What is the potential energy of the r
Anna71 [15]

Answer:

220,500 Joules

Explanation:

PE = (75 kg) x (9.8 m/s^{2}) x (300 m)

PE = (735) x (300)

PE = 220,500 Joules

6 0
3 years ago
Read 2 more answers
What is the kinetic energy of an object that has a mass of 50.0kg and a velocity of 18 m/s?
Rom4ik [11]
<h2>Answer </h2>

The kinetic energy is 8100 J.

<u>Explanation</u>

Mass is 50.0kg and velocity is 18 m/s, the kinetic energy is:

As we know the formula of kinetic energy which is K.E = ½ ( mv ^ 2 ),

mass = m = 50.0kg

velocity = v = 18 m / s,

by putting values in the formula,

K.E = ½ ( mv ^ 2 ),

K.E = ½ ( 50kg ) . ( 18 m / s ) ^ 2

K.E = ½ ( 50kg ) . ( 324 ),

=> K.E = 1/2 ( 16200 ),

=> K.E = 16200 / 2,

=> K.E = 8100J.

Hence, the kinetic energy ( K.E ) is 8100 joule ( J ).  

7 0
3 years ago
If a pebble is being transported in a stream by rolling, how does the velocity of it compare to the velocity of the stream?
AleksandrR [38]
Streams carry sediment, like pebbles, in their flows. The pebbles can be in a variety of locations in the flow, depending on it's size, the balance between the upwards velocity on the pebble (drag and lift forces), and it's settling velocity.
3 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
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