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3 years ago

A 5.0-kg bag of groceries is tossed onto a table at 2.0 m/s and slides to a stop in 1.6 s .

Physics

1 answer:

son4ous [18]3 years ago

5 0

Answer:

-6.3 N

Explanation:

The equation we need to use is:

$F \Delta t = \Delta (mv)$

where

F is the force of friction, that slows down the bag

$\Delta t$ is the time of the motion

m is the mass of the bag

v is the speed of the bag

Since the mass of the bag does not change, we can rewrite the equation as

$F \Delta t= m \Delta v$

where we have:

$\Delta t=1.6 s\\m=5.0 kg\\\Delta v=-2.0 m/s$

Substituting and re-arranging the equation, we can find the force of friction:

$F=\frac{m\Delta v}{\Delta t}=\frac{(5.0 kg)(-2.0 m/s)}{1.6 s}=-6.3 N$

where the negative sign means that the force is in the opposite direction to the motion of the bag.

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3 years ago

You are lost at night in a large, open field. Your GPS tells you that you are 122.0 m from your truck, in a direction 58.0o east

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Answer:

The person is 187[m] farther and 70° south to east.

Explanation:

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We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.

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4 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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3 years ago

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lana [24]

(a) Let's convert the final speed of the car in m/s:
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So we need to find the acceleration first:
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3 years ago

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MatroZZZ [7]

Solid elements are rigid elements. For example an element iron is in solid form. You can touch it and it’s hard.

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