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natima [27]
2 years ago
10

A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.

Physics
1 answer:
Inessa [10]2 years ago
6 0

Hi there!

We can use Newton's Second Law:
\Sigma F = ma

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}

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Two loudspeakers 42.0 m apart and facing each other emit identical 115 Hz sinusoidal sound waves in a room where the sound speed
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Answer:

She passes through a loud spot at x = 19.5m from the first speaker. Constructive interference occurs here.

At the quiet spot destructive interference occurs. The minimum distance for this to occur is 20.25 from the first speaker.

Explanation:

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Which scenario did not include a chemical change?
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3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

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2 years ago
Is directly proportional to the length and resistivity of the conductor.
aliya0001 [1]

Answer:

B) resistance

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