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kiruha [24]
3 years ago
12

I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?

Physics
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

Δy = 68.5

The egg fell 68.5 meters.

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As Clinton walks he pushes his shoe against the track.
Valentin [98]

Answer:A:The track pushes back on Clinton's shoe with the same force.

Explanation:According to Newton's third law of motion, for every action force there is an equal and opposite reaction force. In this case, the action force is Clinton's shoe pushing on the track. As this happens, there is an equal and opposite reaction force in which the track pushes back on Clinton's shoe with the same force.

4 0
3 years ago
Which is an example of gaining a static charge by conduction?
lianna [129]

Answer:

shuffling shoes

Explanation:

7 0
3 years ago
Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col
Ratling [72]
In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
p_f=(m+M)v_f=4.8 kg m/s
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find v_f, the velocity of the nail and the block after the collision:
v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}=  0.48 m/s

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
p_i = mv_i = p_f
Rearranging the formula, we can find v_i, the velocity of the nail before the collision:
v_i =  \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s
6 0
3 years ago
Read 2 more answers
A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

  • u = 8m/ s
  • \theta = 40°

calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

=  \frac{64 \times  \sin(80) }{10}

value of sin 80 = 0. 985

=  \frac{64 \times 0.985}{10}

=  \frac{63.027}{10}

= 6.3027

Hence,

\mathfrak { \blue{the \: teammate \: is \:  \red{\underline{6.3027 \: meters} }\: away } }

7 0
3 years ago
A small 25 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe’s kinetic energy?
Helen [10]
Kinetic energy = 1/2mv^2

=1/2(25)(1^2)
= 12.5J
4 0
3 years ago
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