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AfilCa [17]
3 years ago
6

A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp

osite direction with a force of 20 N. What is the acceleration of the object?
5 m/s2, left

5 m/s2, right

20 m/s2, left

20 m/s2, right
Physics
1 answer:
alex41 [277]3 years ago
5 0

Answer:

5 m/s2, left

Explanation:

We can solve the problem by applying Newton's second law of motion, which  states that:

\sum F=ma

where:

\sum F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, we have:

\sum F=30 N - 20 N = 10 N (to the left) is the net force on the object

m = 2.0 kg is the mass

So, the acceleration is:

a=\frac{\sum F}{m}=\frac{10}{2.0}=5.0 m/s^2

in the same direction as the force (left).

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What would be the velocity<br>when a dog of 10kg and it's kinetic energy is 20J​
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If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti
Ann [662]
We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km
8 0
3 years ago
calcula la potencia por hora de un radiador, sabiendo que esta conectado a un contacto común 110 v. y requiere 20 Amp.
Ira Lisetskai [31]

calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.

Answer:

2.2kWh

Explanation:

Given parameters:

Potential difference  = 110v

Current  = 20A

Unknown:

Power  = ?

Solution:

To solve this problem, we use the expression below:

        Power  = IV

 Power  = 110 x 20  = 2200W

  This is therefore 2.2kW

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8 0
3 years ago
The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10
loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

3 0
2 years ago
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