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3 years ago

If santa slides 5.00 m before reaching the edge, what is his speed as he leaves the roof?

1 answer:

8 0

Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.

Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is

Square root of (19.6h) meters per second.

It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.

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The length of the track itself is 500m. The vehicles (each of mass 15kg) are dragged

gogolik [260]

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

$W = Fd$

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

$W = (200\ N)(500\ m)\\$

5 0

3 years ago

A bowling ball accidentally falls out of the cargo bay of an airliner as it flies along in a horizontal direction. As seen from

valina [46]

The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

<h3>Path of the bowling ball</h3>

Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.

Thus, the path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

Learn more about horizontal direction here: brainly.com/question/2534565

#SPJ1

3 0

2 years ago

1. The following can be inferred from Newton’s second law of motion except:

lora16 [44]

Answer: 1.d) The acceleration of an object is always less than the acceleration due to gravity, g (9.81m/s^-2)

2.a)acceleration decreases

Explanation:

Newton's second law:

Newton's second law states that the acceleration of an object is defined by two variables which is the total force acting on the object and the mass of that object. The acceleration is directly proportional to the net force that is applied on an object and inversely proportional to the mass of that object.

When the force applied on an object is increased so does the acceleration of an object however if the mass increase the acceleration decreases.

This can be felt when you look at the truck which usually carry heavy loads they seem to drive slow due to the load hence their acceleration is decreased by the mass that these truck carry .

7 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

3 years ago