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Romashka-Z-Leto [24]
3 years ago
9

An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc

ean liner from a height of 5.5 meters above the ferryboat’s deck. Calculate the distance at which the ferryman should position the boat from the point horizontally below the point of release so that the parcel lands inside the boat.
Physics
1 answer:
Afina-wow [57]3 years ago
4 0
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

s = ut + 1/2 at²

where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity. 

5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds

Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time

The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

Distance = 10 * 1.06
Distance = 10.6 meters

The boat should be 10.6 meters away horizontally from the point of release.
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Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti
Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

mv_{1i}+Mv_{2i}=(m+M)v  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

3 0
2 years ago
Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl
lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

   m g sin\theta - F_{rope} = ma

 velocity is constant, a = 0

   m g sin\theta - F_{rope} =0

   F_{rope} = m g sin\theta

   F_{rope} = 70.1\times 9.8\times sin 8.6^0

  F_{rope}= 102.73\ N

b) now, when acceleration, a = 0.135 m/s²

   F_{rope}-m g sin\theta = ma

 velocity is constant, a = 0.135 m/s₂

   F_{rope} = m g sin\theta+ma

   F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135

  F_{rope}= 112.19\ N

7 0
3 years ago
What formula should I use?
Elodia [21]
These nuts on your chin
7 0
2 years ago
The speed of a car during a trip is represented in the graph. Compare distance and displacement at four hours into the trip.
mezya [45]

Answer:

The answer to your question is: D.

Explanation:

Distance refers to the amount of space between two points, it is a scalar quantity.

Displacement refers to the space between two points, measure from the minimum path linking them, it is a vector quantity.

I'm not agree with these answers, because the total distance  is approximately 500km.

A) The distance traveled is 300 km.  This answer is not correct.

B) Distance is 300 km and displacement is 0 km.  This answer is not correct because the displacement is also 500 km.

C) Distance is 300 km/hour and displacement is 300 km. 300 km/h is a measure of speed not of distance, this option is wrong.

D) Both distance traveled and displacement are 300 km. I think this is the correct answer because distance and displacement measure the same. but I think both measure 500 km.

7 0
3 years ago
Read 2 more answers
65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!
otez555 [7]
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.



3 0
3 years ago
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