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Ulleksa [173]
3 years ago
10

An 80.0 kg hiker walks a distance of 400.0 m along a road that slopes 5.0 degrees upward, and then stops. What is the hiker's fi

nal gravitational potential energy relative to her original position? You may use a trigonometry table
Physics
1 answer:
lana66690 [7]3 years ago
4 0
The height difference is found by
\delta H=400sin(5 \°)=34.86m
Then the change in potential energy is
E=mgh=(80.0kg)(9.8 \frac{kgm}{s^2})(34.86)= 27332J
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A 45 kg merry go round worker stands on the rides platform 6.3 m from the center. If her speed as she goes around the circle is
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I’ve got 120 n, hope it helps!

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The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob
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Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

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2 years ago
At the equator earth rotates with a velocity of about 465 m/s.
Dafna1 [17]
The given velocity is 465 m/s.

Part a.
465 \,  \frac{m}{s} =(465 \times 10^{-3} \,  \frac{km}{s})*( 3600 \,  \frac{s}{h} ) = 1674 \,  \frac{km}{h}
Answer: 1674 km/h

Part b.
1674  \frac{km}{h} = (1674 \,  \frac{km}{h})*(24 \,  \frac{h}{day}  ) = 40176 \,  \frac{km}{day}
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3 0
3 years ago
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What sound signal tells a river lock attendant that you wish to go through the lock?
Hoochie [10]
In order to tell a river lock attendant that you wish to go through the lock, you should <span>sound one prolonged blast followed by one short blast.

You should wait about 400 feet away from the lock and wait for the flashing light signal that allows you to enter.
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6 0
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.

7 0
3 years ago
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