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3 years ago

8

Nikolas, the fire extinguisher, and the skateboard have a combined mass of 50 kg. What force would the fire extinguisher have to

produce to propel Nikolas if he wanted to accelerate at a rate of 1.2 m/s2? Round your answer to the nearest whole newton.

1 answer:

ExtremeBDS [4]3 years ago

6 0

Answer:

60N

Explanation:

Given parameters:

Combined mass = 50kg

Acceleration = 1.2m/s²

Unknown:

Force to propel = ?

Solution:

From Newton's second law of motion, we know that;

Force = mass x acceleration

So;

Force = 50 x 1.2 = 60N

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A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the

Alexus [3.1K]

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end and another test charge at point p

Potential energy is given by = $\frac{k*Q1*Q2 }{R}$

where k is electrostatics constant = $9 *10^9$

Q1 = first charge , Q2= test charge

R= distance between charges

potential at point p

U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = $\sqrt{8^2 + 6^2}$ = 10

New Potential energy

U1 = kq1*q2 / 10

substituting kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

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3 years ago

What does a tachometer measure?

nlexa [21]

A tachometer measures, for example it measures the MPH(Miles Per Hour) in a car. So like it measures your speed, so you could be going 100 MPH.

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4 years ago

Why does the Polaris appear to remain stationary in the night sky when other stars show an apparent pattern?

yanalaym [24]

Because one pole of the Earth's axis of rotation (the North one) points
almost exactly toward Polaris.

If Polaris had a pimple or a bump somewhere on its edge, you'd see
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3 years ago

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proport

netineya [11]

Answer:

$[F]=[MLT^{-2}]$

Explanation:

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :

F = m × a

F is the applied force

m is the mass of the object

a is the acceleration due to gravity

We need to find the dimensions of force. The dimension of force m and a are as follows :

$[m]=[M]$

$[a]=[LT^{-2}]$

So, the dimension of force F is, $[F]=[MLT^{-2}]$ . Hence, this is the required solution.

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3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago