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katen-ka-za [31]
3 years ago
11

two objects were lifted by a machine. One object had a mass of 2 kilograms, was lifted at a speed of 2m/sec . The other had a ma

ss of 4 kilograms and was lifted at 3m/sec. Which object had more kinetic energy while it was being lifted? Show all calculations
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0

Object 2 has more kinetic energy

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

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

where

m is the mass of the object

v is its speed

In this problem, for object 1:

m = 2 kg

v = 2 m/s

So its kinetic energy is

K_1 = \frac{1}{2}(2)(2)^2=4 J

For object 2,

m = 4 kg

v = 3 m/s

So its kinetic energy is

K_2 = \frac{1}{2}(4)(3)^2=18 J

Therefore, object 2 has more kinetic energy.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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Answer:

991.67 miles per day

Explanation:

Since it travels 2975 miles per day, distance traveled in a day =2975/3=991.67 miles

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Explanation:

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3 0
3 years ago
If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k
Paraphin [41]
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
6 0
3 years ago
A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y =
In-s [12.5K]
Position:                 x = 18t    y = 4t - 4.9t²

First derivative:        x' = 18      y' = 4 - 9.8t

Second derivative:    x'' = 0        y'' = - 9.8


Position vector:      P  =  (18t) i  +  (4t - 4.9t²) j

Velocity vector:      V  =  (18) i  +  (4 - 9.8t) j

Acceleration vector  A  =              (- 9.8) j

6 0
3 years ago
Read 2 more answers
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