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MAVERICK [17]
3 years ago
5

What is the net force needed to accelerate a 5 kg object at 3 m/s2? Suppose that in this situation you discovered that there is

a 5N force of friction opposing the motion. How large is the applied force acting on the object?
Physics
1 answer:
Rudik [331]3 years ago
4 0

The Force need to accelerate the object is by 3 m/s² is 15 N.  Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N

<h3>Force:</h3>

This can be defined as the product of the mass and the acceleration of a body. The S.I unit of force is kgm/s or Newton(N)

To calculate the force needed to accelerate a mass of 5 kg object at 3 m/s² we use the formula below.

Formula:

  • F = ma........ equation 1

Where:

  • F = Net force needed to accelerate the object
  • m = mass of the object
  • a = acceleration of the object

From the question,

Given:

  • m = 5 kg
  • a = 3 m/s²

Substitute these values into equation 1

  • F = 5(3)
  • F = 15 N

Suppose a frictional force of 5 N acts on the motion, The force applied is

  • F = F'+ma............ Equation 2

Where:

  • F = Frictional force = 5N

Substitute into equation 2

  • F = 5(3)+5
  • F' = 15+5
  • F = 20 N.

Hence, The Force need to accelerate the object is by 3 m/s² is 15 N. Suppose a friction force of 5 N acts on the motion of the object, the force needed to be applied to the object is 20 N

Learn more about force here: brainly.com/question/12970081

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NO_{1.499}

Explanation:

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n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

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n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

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Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

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3 years ago
Starting from rest and moving in a straight line, a cheetah
UNO [17]

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of  motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2

Thus, the average acceleration of the cheetah is 7.75 m/s²

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A current is flowing in a wire in direction 3i + 4j where the direction of the magnetic field is 5j + 12k. The force on the wire
nasty-shy [4]

Answer:

F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

the correct answer is "b"

Explanation:

The magnetic force is

          F = i l x B

The bold are vectors, in this case they give us the direction of the current and the magnetic field, for which we can solve as a determinant

         

F = i \left[\begin{array}{ccc}x&y&z\\3&4&0\\0&5&12\end{array}\right]

resolver

     F = i ^ (4 12 - 0) + j ^ (0- 3 12) + k ^ (3  5 - 0)

     F = i (48 i ^ - 36 j ^ + 15 k⁾

in this case i is the value of the current flowing through the cable

     i = 16 mA = 0.016 A

     F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

When reviewing the different answers, the correct answer is "b"

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