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ASHA 777 [7]
2 years ago
15

You are stationary on the ground and are watching a bird fly horizontally towards you at a rate of 2 m/s. The bird is located 20

m above your head. How fast does the angle of elevation of your head change when the horizontal distance between you and the bird is 110 m
Physics
1 answer:
Ira Lisetskai [31]2 years ago
3 0

The angle of elevation changes by 0.0032 rad / sec

Let x (t) be the horizontal distance from you to bird.

Let θ be the angle of elevation of your head.

Then .tan  θ = 20 /x

Differentiating

Sec² = dθ / dt = -20/x².dx/dt

dθ / dt = -20/x².dx/dt.cos²θ

dx/dt = -2m/s   x = 110m

cos²θ = 110²/ (110²) + (20²) = 12100/12500

dθ / dt = 20/ (110)² x (-2) . 12100/12500

=  20 x 2/12500 = 0.0032 radian/s.

Hence ,

The angle of elevation changes by 0.0032 rad / sec

Learn more about angle of elevation:brainly.com/question/16716174

#SPJ1

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Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo
777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert  42 m/s ⇒   km/h

1 km =1000 m

1 h = 36000 sec

42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\

0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}

Step 2

find kilometers traveled after 4  hours

V=\frac{s}{t}\\ \\

V,velocity

s, distance traveled

t. time

now, isolating s

V=\frac{s}{t} \\s=V * t\\

and replacing

s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\

S=604.8 Km

Have a great day

4 0
3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
3 years ago
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Tems11 [23]

Answer:

im not sure

Explanation:

5 0
3 years ago
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Fynjy0 [20]

Answer:

its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

3 0
3 years ago
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