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ArbitrLikvidat [17]

3 years ago

8

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t

he velocity at launch?

1 answer:

Lyrx [107]3 years ago

6 0

Given that

Velocity of missile (v) = 20 m/s ,

Angle of missile (Θ) = 53°

Determine , Vertical component = v sin Θ

= 20 sin 53°

= 15.97 m/s

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A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the f

olga nikolaevna [1]

Answer:

$\lambda =533.6 nm$

Explanation:

the slits spacing, d = 0.21 mm

distance of screen, D = 61 cm

The condition for minima is given as

$dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda$

So, first minima, n = 0

$dsin(\theta_1) = \frac{1}{2}\lambda$

fifth minima, n = 4

$dsin(\theta_5) = \frac{9}{2}\lambda$

$d(sin(\theta_5) -sin(\theta_1))= 4\lambda$

For small angle

$d(tan(\theta_5) -tan(\theta_1))= 4\lambda$

From the figure:

$d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda$

$\frac{d}{D} (y_5-y_1) = 4\lambda$

$\lambda = \frac{d}{4D} (y_5-y_1)$

$\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})$

$\lambda =533.6 nm$

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3 years ago

A 30%-efficient car engine accelerates the 1300 kg car from rest to 10 m/s . How much energy is transferred to the engine by bur

WITCHER [35]

Answer:

The Energy transferred to the engine by burning gasoline = <u>216.67 KJ</u>

Explanation:

The parameters given are:

The efficiency of the car engine, E = 30% = 0.3

Mass, m = 1300 kg

Initial velocity, u = 0, since the car is from rest

The final velocity, v = 10 m/s

Since the car was moving, we calculate its kinetic energy.

kinetic energy = ((1/2) (m) (v^2)

((1/2) (1300 kg) (10 m/s^2)

= 65,000 j

The Energy, Q transferred to the engine by burning gasoline in this case

= potential energy / The efficiency of the car engine, E

Q = 65,000 j / 0.3

= 216,666.66 J

Converting Joule to kilojoule

where 1KJ = 1000j

216,666.66 J = <u>216.67 KJ</u>

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3 years ago

The orbital radius of the Earth (from Earth to Sun) is 1.496 x 10^11 m.

mrs_skeptik [129]

Explanation:

The orbital radius of the Earth is $r_1=1.496\times 10^{11}\ m$

The orbital radius of the Mercury is $r_2=5.79 \times 10^{10}\ m$

The orbital radius of the Pluto is $r_3=5.91 \times 10^{12}\ m$

We need to find the time required for light to travel from the Sun to each of the three planets.

(a) For Sun -Earth,

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$T_1^2=\dfrac{4\pi ^2}{GM}r_1^3$

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So,

$T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s$

(b) For Sun -Mercury,

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(c) For Sun-Pluto,

$T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s$

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Why do our eyes see the color red when we look at a tomato?

forsale [732]

Answer:

B. Tomatos reflect red light

Explanation:

The only reason colors exist is because the objects with color reflect all other light except for what they are portrayed as. White reflects all colors, and black absorbs all colors.

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