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sergeinik [125]
3 years ago
15

The mass of a truck with its cargo is 1800 kg. The truck is coasting to the right with a speed of 10 m/s. As it coasts its cargo

of mass 300 kg is lifted up out of the truck by a crane. What will the velocity of the truck be after its cargo is taken off?
Physics
2 answers:
OverLord2011 [107]3 years ago
6 0

Here we apply conservation of linear momentum. The momentum of the truck with cargo and without cargo remains constant. That is,

mv=m'v'.

Here m,v are initial mass and velocity. m',v' are final mass and velocity. Here m=1800,v=10 and m'=1800-300=1500.

The velocity of the truck be after its cargo is taken off is

v'=\frac{mv}{m'} \\  v'=\frac{1800*10}{1500} \\  v'=12 m/s

timofeeve [1]3 years ago
6 0
I would round it out to be 15 ms or 20 good luck 

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please answer 1-3 and use a separate piece of paper thank you i've been trying those problems for 3 day.... ikr Sad
11Alexandr11 [23.1K]
\frac{total distance}{ total time}

\frac{72m}{36s} = 2m/s

Same thing for the rest 
 2) 5km/h
 3)360km/h

4 0
3 years ago
Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe
dusya [7]

Answer:

Acceleration of the ship, a=2.14\times 10^{-7}\ m/s^2

Explanation:

It is given that,

Mass of both ships, m=39000\ metric\ tons=39\times 10^6\ kg

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

F=G\dfrac{m^2}{d^2}

F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

a=\dfrac{F}{m}

a=\dfrac{8.38\ N}{39\times 10^6\ kg}

a=2.14\times 10^{-7}\ m/s^2

So, the acceleration of either ship due to the gravitational attraction of the other is 2.14\times 10^{-7}\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the
san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0
3 years ago
Name and describe the two layers inside Earth that are undergoing convection and indicate one major consequence of each.
Lilit [14]

Answer:

Mantle and core

Explanation:

The Mantle and Core are the two components within Earth experiencing convection. In several ways the mantle is significant. The one outcome of convective current is the creation of the fresh oceanic lithosphere around OCEANIC RIDGES, formed by mantle upwelling. Core is indeed the planet's innermost layer.

5 0
3 years ago
How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'
Serhud [2]

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity (K.E.=\frac{1}{2}mv^2). If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height(P.E.=mgh). If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.


4 0
3 years ago
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