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2 years ago

11

A flying stationary kite is acted on by a force of 9.8 N downward. The wind exerts a force of 45 N at an angle of 50.0 degrees a

bove the horizontal. Find the force that the string exerts on the kite.

1 answer:

Savatey [412]2 years ago

7 0

Answer:

38 N, 40.0° below the horizontal

Explanation:

Force exerted by an object equals mass times acceleration of that object: F = m ⨉ a. To use this formula, you need to use SI units: Newtons for force, kilograms for mass, and meters per second squared for acceleration.

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atroni [7]

3.8 is the smallest number because if you multiple the others would be large

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2 years ago

Why is it incorrect to say heavy objects sink in water?

lesya [120]

It is incorrect to say heavy objects sink in water because based on the density of the water it can actually cause the "heavy object" to float, out weighing it.

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2 years ago

regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

7 0

3 years ago

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

5 0

2 years ago

What is the kinetic energy of a 200 kg satellite as it follows a circular orbit of radius 8x106m around the earth?

motikmotik

Given the equation for the Speed of a Satellite

v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem

we have:

(square root whole term on right side)

v = G Me
———
r

so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)

v = 7055 m/s (which is reasonable)

so utilize the Kinetic Energy Formula

KE = 1/2mv^2

KE = 1/2(200)(7055)^2

KE = 4.977x10^9 J

4 0

3 years ago

Read 2 more answers