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3 years ago

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car

Physics

1 answer:

Novay_Z [31]3 years ago

0 0

Answer:

$4.5 *10^{3}\ kgm/s^{2}$

Explanation:

Given:

$Mass= 1.00 * 10^{3} \ kilograms,\\Acceleration =4.5\ meters/second^{2}$

As we know that

Force=Mass * Acceleration

Putting the value of mass and Acceleration we get ,

$Force\ = \ 1.00 * 10^{3} * 4.5\ m/s^{2}$

$Force\ =4.5 * \ 10^{3} \ kgm/s^{2}$

Therefore Net force is : $4.5 *10^{3}\ kgm/s^{2}$

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3 years ago

If eight water waves pass an ocean buoy each minute, and successive wave crests are 20 m apart, find the wave speed:____________

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Answer:

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A child with mass of 25 kg gets into a toy car with mass of 80 kg on a playground, causing it to sink on its springs (with effec

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3 years ago

The capacitance of A conductor is affected by the presence of A second conductor that is uncharged and isolated electrically. Wh

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3 years ago

An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v

Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is $0.\overline {36}$ C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40 volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = $0.\overline {36}$ C

The charge on the second bulb, Q₂ = $0.\overline {36}$ C.

d. The question has left out parts

4 0

3 years ago