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3 years ago

How much force is needed to accelerate a 1-kilogram toy car at a rate of 2 meters per second per second?

Physics

2 answers:

Feliz [49]3 years ago

7 0

Answer:

2 Newtons

Explanation:

kozerog [31]3 years ago

5 0

F = ma, where m = mass in kg, a = acceleration in m/s², F = Force in Newton

F = 1 * 2

F = 2 N

Force needed is 2 Newtons.

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A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra

Sonja [21]

Answer:

Intensity, $I=1.101\ W/m^2$

Explanation:

Power of the light bulb, P = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

$I=\dfrac{P}{A}$

$I=\dfrac{P}{4\pi r^2}$

$I=\dfrac{40\ W}{4\pi (1.7\ m)^2}$

$I=1.101\ W/m^2$

So, the intensity of light is $1.101\ W/m^2$ .

6 0

3 years ago

A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil cont

EastWind [94]

Answer:

The number of turns in secondary coil is 4000

Explanation:

Given:

Current in primary coil $I_{P} = 500$ A

Current in secondary coil $I_{S} = 25$ A

Number of turns in primary coil $N_{P} = 200$

In case of transformer the relation between current and number of turns is given by,

$\frac{N_{S} }{N_{P} } = \frac{I_{P} }{I_{S} }$

For finding number of turns in secondary coil,

$N_{S} = \frac{I_{P} }{I_{S} } N_{P}$

$N_{S} = \frac{500}{25} \times 200$

$N_{S} = 4000$

Therefore, the number of turns in secondary coil is 4000

5 0

3 years ago

A small charged bead has a mass of 1.9 g . it is held in a uniform electric field e⃗ = (200,000 n/c, up). when the bead is relea

dmitriy555 [2]

As we know that when charge is released in electric field

It will have two forces on it

1. electrostatic force

2. gravitational force

now if the ball will accelerate upwards so we can say

net upward force = mass * acceleration

$qE - mg = ma$

$q*200000 - 1.9 * 10^{-3}* 9.8 = 1.9 * 10^{-3}* 15$

now we can find charge q on it by above equation

$q = 0.2356 * 10^{-6} C$

So above is the charge on the particle

7 0

3 years ago

True of false efficiency compared the output work to the output force

Lerok [7]

The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>

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3 years ago

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3 years ago