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Levart [38]
3 years ago
9

As a hiker in Glacier National Park, you are looking for a way to keep the bears from getting at your supply of food. You find a

campground that is near an outcropping of ice from one of the glaciers. Part of the ice outcropping forms a slope of angle θup to a vertical cliff. You decide that this is an ideal place to hang your food supply, as the cliff is too tall for a bear to reach it. You put all of your food into a burlap sack, tie an unstretchable rope to the sack, and tie another bag full of rocks to the other end of the rope to act as an anchor. The mass of rocks that you put into the anchor bag is equal to the mass of food in the other bag.What will be the acceleration aof the food bag when you let go of the anchor bag? Assume that the weight of the rope is negligible, and that the ice can be considered frictionless.Express the resulting acceleration aof the food bag in terms of θ and g, the acceleration due to gravity. Let the positive direction for this bag be downwards.
Physics
1 answer:
LenKa [72]3 years ago
8 0

Answer: the acceleration is (g( 1 - sinθ)) / 2

Explanation:

the tension in one bag is expressed as;

T = ma + mg sin θ ........................lets say equ 1

now the tension in another bag is expressed as;

T = -ma + mg..................lets say equ 2

so equate both equations

ma + mg sinθ = -ma + mg

2ma =  mg ( 1 - sinθ)

a = (g( 1 - sinθ)) / 2

therefore the acceleration is (g( 1 - sinθ)) / 2

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What instrument is normally used to measure atmospheric pressure?
Anuta_ua [19.1K]
B. Barometer is used to measure atmospheric pressure.
6 0
3 years ago
What do (3.9 x 10^2)(4.1 x 10^4) equal
pantera1 [17]

Answer:

Explanation:(3.9x100)(4.1x10,000)                

390x41,000

15,990,000

8 0
3 years ago
A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle
liq [111]

Answer:

0.0031792338 rad/s

Explanation:

\theta = Angle of elevation

y = Height of balloon

Using trigonometry

tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta

Differentiating with respect to t we get

\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft

Now y = 2500 ft

cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974

\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s

The angle is changing at 0.0031792338 rad/s

6 0
3 years ago
You notice that when you turn off your bedroom lights, the kitchen lights can stay on. Based on this observation, what kind of c
Sonja [21]

Answer:

the correct answer is A

Explanation:

In a series circuit the current flows through the circuit and when it is opened at some point the current stops throughout the circuit.

In a parallel circuit, the current flows through the different branches of the circuit and when the current in one branch is interrupted, it continues to flow through the others.

In this exercises they indicate that the bedroom lights are turned off (the circuit was opened), but the kitchen lights are still on, therefore it is a parallel circuit with at least two branches, the current that circulates through each one depends on the existence on each branch.

Therefore the correct answer is A

3 0
3 years ago
Two spherical objects are separated by a distance that is 1.08 x 10-3 m. The objects are initially electrically neutral and are
Alla [95]

Answer:

the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

Explanation:

from coulumbs law, The force that is acting over both charge can be computed as

F=( kq1q2)/r^2..............eqn(1)

Where

F=electrostatic force= 3.89 x 10-21 N

k= column constant= 9 x 10^9 Nm^2/C^2

q1 and q2= magnitude of the charges

r= distance between the charges= 1.08 x 10-3 m.

Since both charges are experiencing the same force, eqn(1) can be written as

F=( kq^2)/r^2.

We can make q subject of the formula

q= √(Fr^2)/k

= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9

q= 71.043×10^-20 C

Hence, the charge is 71.043×10^-20 C

From quantization law, the number of electron can be computed as

N=q/e

Where

N= number of electron

q= charges

=1.6×10^-19C

N=71.043×10^-20/1.6×10^-19

=4.44

Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

8 0
3 years ago
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