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disa [49]
3 years ago
10

ANSWER 5 AND 8!!!!!;$$4$;$;

Physics
1 answer:
DedPeter [7]3 years ago
4 0

Answer:

5 ) The mass, 6) with lubrication and using surfaces that are not rough

Explanation:

5) If two bodies are held regardless of their densities and can be combined by some chemical or physical process, the only physical property to be modified will be the mass of the resulting body.

8)

Friction depends on the contact between two surfaces and when a body has a relative motion with respect to a contact surface. In order to reduce friction the contact surface must be lubricated, also the friction depends on the coefficient of friction between surfaces and the normal force exerted by the surface parallel to the area of contact with the body. Mathematically it can be expressed with the following equation.

F_{f} = u*N\\where:\\u = friction coefficient\\N = normal force [Newtons]\\F_{f}= friction force [Newtons]

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A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds
ki77a [65]

Answer:

Explanation:

The question is incomplete.

The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.

s= sin2(pi)t

Acceleration = d²S/dt²

dS/dt = 2πcos2πt

d²S/dt² = -4π²sin2πt

A(t) = -4π²sin2πt

Next is to find acceleration after 4.5 seconds

A(4.5) = -4π²sin2π(4.5)

A(4.5) = -4π²sin9π

A(4.5) = -4π²sin1620

A(4.5) = -4π²(0)

A(4.5) = 0m/s²

4 0
3 years ago
Which of the following best explains why most chemical reactions proceed more quickly when the concentrations of reactants are i
nekit [7.7K]
A More concentrated means more collisions per unit volume, and as the volume stays the same and only concentration changes, the there are more collisions
3 0
3 years ago
This type of light production is an example of _______ energy being transformed into light energy.
Anton [14]

Electric energy??? there isn't much information from the question but I can infer it's electricity.

5 0
3 years ago
an acorn falls from a tree. ignoring air resistance, what forces are on the acorn while it’s falling?
Triss [41]

Answer: gravity

Explanation:

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