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fredd [130]
3 years ago
7

6) what is ararge of speed​

Physics
1 answer:
irina [24]3 years ago
5 0

Answer:

in everyday use and in kinematic the speed of an object is the magnitude of the change of its position it is thus a scalar quantity

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Which of the following is the best reason that trees care important for overall air quality?
Marina86 [1]
Trees are important because oxygen
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3 years ago
A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.
Crazy boy [7]
A) Work energy relation;
 Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
 F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
 F = 900/0.04
    = 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
 F = Ma
Thus; a = F/m
             = 22500/(5×10^-3)
             = 4,500,000 m/s²
But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
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Water from a river or reservoir enters a hydroelectric power plant through a.....????
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Which of the following situations is an example of kinetic energy being transformed into potential energy? Throwing a ball high
motikmotik

Answer: holding the ball in the air without moving it

Explanation:

When the ball goes up it’s potential energy

6 0
2 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
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