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jek_recluse [69]
3 years ago
11

While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se

cond stone straight down. what initial velocity must you give the second stone if they are both to reach the ground at the same instant? take the downward direction to be the negative direction?
Physics
1 answer:
Norma-Jean [14]3 years ago
3 0
Given: distance 1 d₁ = 40 m;  distance 2 d₂ = 3.8 m   g = -9.8 m/s²

 Initial Velocity Vi = 0  Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂  = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² -  Vi²/2d   derive for Final Velocity Vf
 
acceleration is now due to gravity, therefore a = g

Vf = √2gd   Vf = √2(9.8 m/s²)(36.2 m)

Vf = 26.64 m/s  

Reason: The second stone will still start from rest.


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Temperature which is solid becomes a liquid
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The answer would be melting point.

Explanation:

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You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo
AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

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4 0
2 years ago
A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele
Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

3 0
3 years ago
A forklift pushes a box with a force of 500 N across the floor for a distance of 5.0 m, then turns around and pushes with the sa
yanalaym [24]

Answer:

5,000J

Explanation:

Work = Force x Distance

Distance back and forth is canceled out, so either the answer is + or -

5.0m + 5.0m = 10.0m

500N x 10.0m = 5,000J

4 0
3 years ago
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