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Alexxx [7]
3 years ago
8

Which object would have MORE kinetic energy?

Physics
2 answers:
GuDViN [60]3 years ago
7 0

Answer:

energy \:  =  \frac{1}{2} m {c}^{2}  \\ energy \: is \: directly  \\ \: protional \: to \: mass \\ the \: hevier \: the \: object \\ the \: more \: k.e.

valentinak56 [21]3 years ago
6 0

Answer:

A dump truck going 70 mph

Explanation:

hope it helps you

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Find the change in kinetic energy of a 0.650 kg fish leaping to the right at 15.0 m/s that collides inelastically with a 0.950 k
yaroslaw [1]
The change in the kinetic energy refers to the work done in displacing a body, thus, the change in the kinetic energy of an object refers to the work done on the object.
The correct formula to use is:
W = Initial kinetic energy - Final kinetic energy;
Where, W = change in kinetic energy
Final kinetic energy and initial kinetic energy = 1/2 MV^2
Initial velocity = 15 m/s
Final velocity  = 13.5 m/s
Initial mass = 0.650 kg
Final mass = 0.950 kg
W = 1/2 [0.650* (15 *15)] - 1/2 [0.950 * (13.5 * 13.5)]
W = 146.25 - 173.13 = 26.88
Therefore, the change in kinetic energy is 26.88 J.
 The negative sign has to be ignored, because change in kinetic energy can not be negative.
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4 0
3 years ago
Read 2 more answers
A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5
svlad2 [7]
The magnetic force experienced by the proton is given by
F=qvB \sin \theta
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and \theta the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so \sin \theta=1 and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
F=ma

So we have
ma=qvB
from which we can find the magnitude of the field:
B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T
4 0
3 years ago
How are Earth and Venus similar? How is Venus different from Earth? (Provide two similarities and two differences.)
Alika [10]
Venus shares a similar size, surface composition, and has an atmosphere with a complex weather system. Venus is different from Earth because it spins the opposite direction of Earth and it’s rotation is very slow.
8 0
3 years ago
Studen
Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

  D =   72 \  m

b

Magnitude

             d =  - 20 \ m

 Direction  

West

Explanation:

From the question we are told that

The first distance covered westward is d_w_1  =  18 \  m /tex]     The  first distance covered eastward is [tex]d_e1 =  12 \  m /tex]      The second distance covered westward is [tex]d_w_2  =  28 \  m /tex]      The  second distance covered eastward is [tex]d_e2 =  14 \  m /tex]   Generally the distance covered is mathematically represented as       [tex]D =  d_w1 + d_w2 + d_e1 + d_e2

=> D =   18 + 28 + 12 +  14

=> D =   72 \  m

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

d =  -d_w1 -d_w2 + d_e1 + d_e2

=>  d =  -18-28 + 12 + 14

=> d =  - 20 \ m

The direction is west

7 0
3 years ago
what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
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