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Gnesinka [82]
3 years ago
13

Flying against the wind, an airplane travels 2670 km in 3 hours. Flying with the wind, the same plane travels 11,070 km in 9 hou

rs. What is the rate of the plane in still air and what is the rate of the wind?
Physics
1 answer:
xxMikexx [17]3 years ago
5 0

Answer:

speed of plane in still air = 1060 km/h

speed of wind = 170 km/h

Explanation:

Let teh speed of plane in still air is vp and the speed of air is va.

Irt travels 2670 km in 3 hours against the wind

So,

vp - va = 2670 / 3 = 890 km/h ..... (1)

It travels 11070 km in 9 hours along the wind.

vp + va = 11070 / 9 = 1230 km/h .... (2)

Adding both the equations

2 vp = 2120

vp = 1060 km/h

and va = 1230 - vp = 1230 - 1060 = 170 km/h

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A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c
Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}

hence, the speed of the spheres is 37.45 m/s

8 0
3 years ago
A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the
Alexus [3.1K]

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end  and another test charge at point p

Potential energy is given by  = \frac{k*Q1*Q2 }{R}

where k is electrostatics constant = 9 *10^9

Q1 = first charge , Q2=  test charge

R= distance between charges

potential at point p

U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = \sqrt{8^2 + 6^2} = 10

New Potential energy

U1 = kq1*q2 / 10

substituting  kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

5 0
3 years ago
If the speed of a particle is doubled, what happens to its kinetic energy?
Norma-Jean [14]

Answer:

it increases

Explanation:

maybe for example we have 5pets then you multiple them by two u get 10

4 0
3 years ago
In a chemical reaction blank are the substances left over
Zepler [3.9K]

Answer: In a chemical reaction blank are the substances left over

Explanation: The substance left over after a reaction takes place

3 0
3 years ago
A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a
marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2.

where y(t) represent the height from the ground. For our problem, the initial height will be:

y_0 \ = \ 1000 m.

The initial velocity:

v_0 = - 20 \frac{m}{s},

take into consideration the minus sign, that appears cause the ball its thrown down.  The same minus appears for the acceleration:

a=-10\frac{m}{s}

So, the equation for our problem its:

y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2.

Taking t=6 s:

y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2.

y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2.

y(6 \ s) = \ 1000 m \ - 120 m - 180 m.

y(6 \ s) = \ 1000 m \ - 300 m.

y(6 \ s) = \ 700 m.

So this its the height of the ball 6 seconds after being thrown.

6 0
3 years ago
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