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myrzilka [38]
3 years ago
7

Consider a person sliding down a water slide at constant velocity. what are the forces acting on the person as they slide? are t

he forces balanced or unbalanced? explain.

Physics
2 answers:
Andru [333]3 years ago
4 0

Explanation :

When a person is sliding with constant velocity, the forces that are acting on him are :

(1) Weight, W = mg where g is acceleration due to gravity.

(2) Friction forces acting opposite to the direction of motion.

(3) Normal force.

The forces acting on the person is balanced because the person is sliding with constant velocity.

From figure it is clear that, friction force is balanced by w\ sin\theta and normal is balanced by w\ cos\theta.

Umnica [9.8K]3 years ago
4 0

Answer:

It is D.

Explanation:

Gravity and friction are acting on the person. The forces are balanced. There is no net force because the person is moving at a constant velocity. In this example, there is no net force when a mass moves at constant velocity. Although friction is acting on the person, there is no change in velocity and friction is not a net force in this case. Friction is only a net force if it changes the velocity of a mass. If friction slowed the person down, then there is a net force.

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What does hydraulic mean?
Harrizon [31]

Answer:

denoting, relating to, or operated by a liquid moving in a confined space under pressure.

4 0
3 years ago
A 4 kg bowling ball moving at 1.4 m/s east impacts a 400 g pin that is stationary. After the impact, the ball is moving at 0.5 m
nignag [31]

The speed of the pin after the elastic collision is 9 m/s east.

<h3>Final speed of the pin</h3>

The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;

m1u1 + mu2 = m1v1 + m2v2

where;

  • m is the mass of the objects
  • u is the initial speed of the objects
  • v is the final speed of the objects

4(1.4) + 0.4(0) = 4(0.5) + 0.4v2

5.6 = 2 + 0.4v2

5.6 - 2 = 0.4v2

3.6 = 0.4v2

v2 = 3.6/0.4

v2 = 9 m/s

Thus, The speed of the pin after the elastic collision is 9 m/s east.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

3 0
1 year ago
A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?
Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒  Density = \frac{Mass}{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

On substituting the values, we get

⇒               =\frac{6.3 \ g}{19.3 \ g/mL}

⇒               =0.32 \ mL

7 0
3 years ago
Relative motion can best be defined as
geniusboy [140]
Relative motion can best be defined as B<span> the motion of one object as it appears to another object.
An example is when you are in a car the car has the actual motion because it is the one moving but you are also moving because of relative motion.</span>
6 0
3 years ago
A stuntman of mass 48 kg is to be launched horizontally out of a spring-
Damm [24]

The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s.  The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

  • mv²/2 = ke²/2
  • mv² = ke².................. Equation 1

⇒ Where:

  • m = mass of the stuntman
  • v = velocity of the stuntman
  • k = force constant of the spring
  • e = compression of the spring

⇒ Make v the subject of the equation

  • v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

  • m = 48 kg
  • k = 75 N/m
  • e = 4 m

⇒ Substitute these values into equation 2

  • v = √[(75×4²)/48]
  • v = √25
  • v = 5 m/s.

Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: brainly.com/question/10962624

6 0
2 years ago
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