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Alinara [238K]
2 years ago
7

A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic

opter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?
Physics
1 answer:
andreev551 [17]2 years ago
8 0

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

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A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the gro
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Answer:

torque is 1.7 * 10^{-2} Nm

Explanation:

Given data

turns n = 1000 turns

radius r  = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

torque is 1.7 * 10^{-2} Nm

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2 years ago
A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is
tiny-mole [99]

Answer:

F=mg(sin(\theta )-0.25 cos(\theta ))

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\

Using value of N from equation β in α we get value of force as

F=mg(sin(\theta )-\mu cos(\theta ))

Applying values we get

F=mg(sin(\theta )-0.25 cos(\theta ))

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1 year ago
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If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
Sav [38]
<h2>Complete Question:</h2>

You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.60 kΩ .

(a) If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?

(b) How much electrical power is delivered to your body?

<h2>Answer:</h2>

(a) 0.075A

(b) 9W

<h2>Explanation:</h2>

The voltage (V) passing across or supplied to a body is directly proportional to the current (I) flowing through the body as stated by Ohm's law. i.e

V ∝ I

=> V = I x R                 ----------------------(i)

Where;

R = constant of proportionality called resistance of the body

(a) As stated in the question;

The body is wet and thus will conduct electricity and has the following;

V = voltage supplied = 120V

R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω

Substitute these values into equation(i) as follows;

120 = I x 1600

Solve for I;

I = \frac{120}{1600}

I = 0.075A

Therefore the amount of current that will pass through your body is 0.075A

(b) Electrical power(P), which is commonly measured in Watts(W), delivered to a body is the product of the current(I) and voltage (V) supplied to the body. i.e

P = I x V           ---------------------(ii)

Where;

I = 0.075A   [as calculated above]

V = 120V     [given in the question]

Substitute these values into equation (ii) as follows;

P = 0.075 x 120

P = 9W

Therefore, the electric power delivered to your body is 9W

7 0
1 year ago
A windowpane is half a centimeter thick and has an area of 1.0 m2. The temperature difference between the inside and outside sur
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L = 5*10^{-3}m Length

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Replacing with our values we have that,

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\frac{Q}{t} = \frac{(0.84)(1)}{0.005} (15)

\frac{Q}{t} = 2520J/s

Therefore the correct answer is B.

3 0
1 year ago
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