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san4es73 [151]
3 years ago
5

A fire helicopter carries a 580-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

d of 40.0 m/s, the cable makes an angle of 38.0° with respect to the vertical. Determine the force exerted by air resistance on the bucket.
Physics
1 answer:
inn [45]3 years ago
4 0

Answer:

F = 41,954 N

Explanation:

given,

mass of bucket = 580 Kg

length of the cable = 20 m

velocity = 40 m/s

angle made = 38.0°

T cos 38° = m g..............(1)

T sin 38^0 = \dfrac{mv^2}{l} + F......(2)

dividing equation (2) by (1)

tan 38^0 = \dfrac{\dfrac{mv^2}{l} + F}{mg}

tan 38^0 = \dfrac{\dfrac{580\times 40^2}{20} + F}{580 \times 9.81}

4445.36 = \dfrac{580\times 40^2}{20} + F

F = -46400 + 4445.36

F = -41,954 N

hence, the force is acting in the opposite direction as assumed.

F = 41,954 N

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A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s
PolarNik [594]

Answer:

The value of radiation pressure is 2.933 \times 10^{-8} Pa

Explanation:

Given:

Intensity I = 8.8 \frac{W}{m^{2} }

Area of piece A = 2.1 \times 10^{-4} m^{2}

From the formula of radiation pressure in terms of intensity,

   P = \frac{I}{c}

Where P = radiation pressure, c = speed of light

We know value of speed of light,

 c = 3 \times 10^{8} \frac{m}{s}

Put all values in above equation,

  P = \frac{8.8}{3 \times 10^{8} }

  P = 2.933 \times 10^{-8} Pa

Therefore, the value of radiation pressure is 2.933 \times 10^{-8} Pa

8 0
3 years ago
Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay
kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

mu=(m+M)v

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s

8 0
3 years ago
A car moving at a speed of 36 km/h reaches the foot of a smooth
boyakko [2]

Answer:

d = 10.2 m

Explanation:

When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

Kinetic\ Energy\ of\ the \ Car = Work\ Done\ while\ moving\ up\ the\ plane\\\frac{1}{2}mv^{2} = Fd

where,

m = mass of car

v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)

v = 10 m/s

F = force on the car in direction of inclination = W Sin θ

W = weight of car = mg

θ = Angle of inclinition = 30°

d = distance covered up the ramp = ?

Therefore,

\frac{1}{2}mv^{2} = mgdSin\theta\\\frac{1}{2}v^{2} = gdSin\theta\\\frac{1}{2}(10\ m/s)^{2} = d(9.81\ m/s^{2}) Sin\ 30^{0}

<u>d = 10.2 m</u>

4 0
2 years ago
!! 50 points spent, easy points for you. 8th-grade science/physics !!
luda_lava [24]

Answer:

1.mirage

2.lens

3.concave lens

4.convex lens

5.index of refraction

Explanation:

IQ=999999

7 0
2 years ago
HELP PLEASE
4vir4ik [10]

Recall this kinematic equation:

a = \frac{Vi+Vf}{Δt}

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s


Substitute the terms in the equation with the given values and solve for a:

a = \frac{0+25}{10}

<h3>a = 2.5 m/s²</h3>
7 0
3 years ago
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