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seropon [69]
3 years ago
10

As an airplane is flying 20.0 degrees north of east at a speed of 750 km/h. How fast is it moving to the north

Physics
2 answers:
dedylja [7]3 years ago
7 0
<span>in this case velocity of airoplane will be 750*sin20'


</span>
Gwar [14]3 years ago
3 0

Answer:

v_y=71.25\ m/s

Explanation:

Given that,

Speed of an airplane, v = 750 km/hr = 208.33 m/s

An airplane is flying 20.0 degrees north of east, \theta=20^{\circ}

To find,

Velocity in north direction.

Solution,

Let v_y is the speed of the airplane to the north. The y component of any motion is given by :

v_y=v\ sin\theta

v_y=208.33\ m/s\ sin(20)

v_y=71.25\ m/s

So, the airplane is moving with a speed of 71.25 m/s to the north direction.

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Se coloca agua en un recipiente de aluminio y se pone a calentar en una estufa que le suministra 230 kj, lo cual hace que la tem
tensa zangetsu [6.8K]

Answer:

32 °C.

Explanation:

Hola.

En este caso, debemos entender que la relación entre el calor y la temperatura viene dada por:

Q=mCp\Delta T

De este modo, dado que estamos estudiando la misma sustancia (agua) con masa constante, la relación calor-temperatura es lineal y directamente proporcional, por tal razón, si se duplica el calor suministrado, la temperatura también será duplicada, de modo que:

\Delta T_{nuevo}=2*16\°C\\\\\Delta T_{nuevo}=32\°C

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3 0
3 years ago
Examine the lightbulbs in the circuit below. Write a sentence explaining what would happen if lightbulb A burned out. Repeat thi
tatiyna

If the lightbulb A in the circuit shown in the image burned out, the path for the  current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.

If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.

8 0
3 years ago
A rifle fires a bullet at a velocity of 78 m/s, 40 degrees above the horizontal. Determine the height (h) above the starting pos
qwelly [4]

Answer:

H = Vy t - 1/2 g t^2  height of an object with an initial "vertical" velocity

                                 at t sec after firing

Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s

H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m

7 0
3 years ago
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i
Rashid [163]

Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = N_{p} = 500 turns

input voltage = V_{p} = 120 V

number of secondary turns = N_{s} = 3 turns

output voltage = V_{s} = ?

using the equation for a transformer

\frac{V_{s} }{V_{p} }  = \frac{N_{s} }{N_{p} }

substituting values, we have

\frac{V_{s} }{120 }  = \frac{3 }{500} }

500V_{p}  = 120*3\\500V_{p} = 360

V_{p} = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

I_{P}V_{p} = I_{s}V_{s}

where

I_{p} = input current = ?

I_{s} = output voltage = 3.2 A

V_{s} = output voltage = 0.72 V

V_{p} = input voltage = 120 V

substituting values, we have

120I_{p} = 3.2 x 0.72

120I_{p} = 2.304

I_{p}  = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = I_{p} V_{p}

==> 0.0192 x 120 = <em>2.304 Watts</em>

7 0
3 years ago
you slide across home plate during baseball game. If you have a mass of 82 kg, and the coefficient of kinetic friction between y
Tju [1.3M]

m = mass of the person = 82 kg

g = acceleration due to gravity acting on the person = 9.8 m/s²

F = normal force by the surface on the person

f = kinetic frictional force acting on the person by the surface

μ = Coefficient of kinetic friction = 0.45

The normal force by the surface in upward direction balances the weight of the person in down direction , hence

F = mg                                          eq-1

kinetic frictional force on the person acting is given as

f = μ F

using eq-1

f = μ mg

inserting the values

f = (0.45) (82) (9.8)

f = 361.6 N

8 0
3 years ago
Read 2 more answers
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