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dolphi86 [110]
3 years ago
5

Hi guys this is my last points and i need some answers :)

Physics
1 answer:
Zinaida [17]3 years ago
6 0

Answer:

Number 1: B  Number 2: D

Explanation:

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The work done on an elevator that has a mass of 1000kg and lifted 4m in 10 seconds is how many joules.
GaryK [48]

Answer:

39,200 Joules

3920 watts

Explanation:

5 0
4 years ago
Which example involves the transformation of chemical energy directly into light energy?
postnew [5]
B) a glow stick , glow sticks are made up of certain chemicals and when they fuse together they start to glow which would be considered as light energy
6 0
3 years ago
Read 2 more answers
When a force is applied to a wheel, its axle exerts a greater force?
makvit [3.9K]

Answer:

That is true.

Explanation:

When the input force is applied to the wheel, as it is with a doorknob, the axle turns over a shorter distance but with greater force, so the mechanical advantage is greater than 1.

7 0
3 years ago
Consider a rifle, which has a mass of 2.44 kg and a bullet which has a mass of 150 grams and is loaded in the firing chamber.Whe
svetlana [45]

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N

According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:

W = K_e

dF_f = 0.5Mv_0^2

d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m

j.Kinetic energy of the bullet before the impact:

K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J

Kinetic energy of the block-bullet system after the impact:

K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J

So 14520 - 33 = 14487 J was lost during the lodging process.

3 0
3 years ago
A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal
sukhopar [10]

Answer:

-6112.26  J

Explanation:

The initial kinetic energy, KE_i is given by

KE_i=0.5mv_1^{2} where m is the mass of a body and v_i is the initial velocity

The final kinetic energy, KE_f is given by

KE_f=0.5mv_f^{2} where v_f is the final velocity

Change in kinetic energy, \triangle KE is given by

\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for v_f and 12.6 m/s for v_i and 77 Kg for m we obtain

\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26  J</u>

3 0
3 years ago
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