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MAVERICK [17]
2 years ago
8

A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?

Physics
1 answer:
Vladimir [108]2 years ago
7 0

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

F=ma\\\\a=\frac{F}{m}

where,

  • a = acceleration = ?
  • F = Magnitude of the applied force = 6 N
  • m = Mass of the ball = 9 kg

Therefore,

a=\frac{6\ N}{9\ kg}

a = 0.67 m/s²

Learn more about Newton's Second Law of Motion here:

brainly.com/question/13447525

#SPJ1

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Well the similarity is that even though they are in a different state of matter they still come from the same substance: h2o 
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3 years ago
I need help with this
mojhsa [17]

Answer:

1) distance walked = 210 m

2) not sure

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4) For avg vel, use the formula - displacement/time

Explanation:

im not fully sure but i studied this last year

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3 years ago
From a lake, water is pumped at a rate of 67 L/s to a storage tank positioned 14 m above while consuming 16.4 kW of electrical p
LenaWriter [7]

Answer:

57 %

Explanation:

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Mass of water pumped per second, m = Volume of water pumped epr second x density of water

m = 67 x 10^-3 x 1000 = 67 kg/s

height raised, h = 14 m

Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W

efficiency = output power / input power = 9380 / 16400 = 0.57

% efficiency = 57 %

thus, the efficiency of the pump is 57 %.

3 0
3 years ago
Robbie found a rock in a stream. The rock was smooth and round. What MOST LIKELY caused this to happen?
Degger [83]

Answer:

C, weathering by the water.

Explanation:

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3 years ago
A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

M = 16.67Kg

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0
3 years ago
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