Which range of the pH scale would most likely include acid rain?

A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total

NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is $64.59^{\circ}$ South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is $64.59^{\circ}$ South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, $\vec{AB} = 446 km$

Distance moved from B to C in South direction, $\vec{BC} = - 939 km$

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(446)^{2} + (- 939)^{2}}$ = 1039 km

(b) Direction of the displacement is given by:

$tan\theta = \frac{\vec{BC}}{\vec{AB}}$

$\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}$

$64.59^{\circ}$ South of East

(c) Magnitude of the average speed is given by:

$v_{avg} = \frac{AC}{t + t'}$

$v_{avg} = \frac{1039}{1.82} = 570.88 km/h$

(d) The direction of the average velocity is the same as that of the displacement, i.e., $64.59^{\circ}$ South of East.

(e) The average speed of the [plane is given by:

$v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}$

$v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h$

6 0

3 years ago

¡¡first convert revolutions per second to

katovenus [111]

Answer:

a. 5.23 m/s² b. 44.23 N

Explanation:

a. What is the centripetal acceleration of the hammer?

The centripetal acceleration a = rω² where r = radius of circle and ω = angular speed.

Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.

So a = rω²

= 1.4 m × (3.74 rad/s)²

= 5.23 m/s²

b. What is the tension in the chain?

The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.

So, T = 8.45 kg × 5.23 m/s²

= 44.23 N

6 0

3 years ago

Please help. what's the answer??

Anvisha [2.4K]

I THINK it is C. KEYWORD is THINK

5 0

3 years ago

What do astronomers expect will eventually happen to the Milky Way galaxy?

Nataly [62]

The Andromeda Milkyway is a galactic collision predicted to occur in 4 billion years. I believe thats the answer

5 0

3 years ago

Read 2 more answers

It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion.

natali 33 [55]

Answer:

The ratio of gravitational force to electrical force is 3.19 x 10^-36

Explanation:

mass of an alpha particle = 6.64 x $10^{-27}$ kg

charge on an alpha particle = +2e = +2(1.6 x $10^{-19}$ C) = 3.2 x $10^{-19}$ C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = $\frac{Gm^{2} }{r^{2} }$

where G = gravitational constant = 6.67 x $10^{-11}$ m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = $\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }$ = $\frac{2.94*10^{-63} }{d^{2} }$ Newton

For electrical repulsion:

Electrical force between the particles = $\frac{-kQ^{2} }{r^{2} }$

where k is the Coulomb's constant = 9.0 x $10^{9}$ N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = $\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }$ = $\frac{-9.216*10^{-28} }{d^{2} }$ Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = $\frac{2.94*10^{-63} }{9.216*10^{-28} }$

==> 3.19 x 10^-36

8 0

3 years ago