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3 years ago

Can someone please help me

Physics

1 answer:

VikaD [51]3 years ago

4 0

Answer:

I don't know the answer but I needed the answer to that on a quiz and I downloaded sorcatic and it brings u to an app or website with the answer I hope this helps if you can't find the app than just tell me

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A sound wave is a

Stolb23 [73]

A sound wave is a longitudinal wave

6 0

3 years ago

A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.

Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light $=15 feet$

height of man $=6 feet$

$\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s$

From diagram

$\frac{15}{y}=\frac{6}{y-x}$

$5(y-x)=2y$

$3y=5x$

differentiate both sides

$3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Tip of shadow is moving at the rate of

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s$

(b)rate at which length of his shadow is changing

Length of shadow is $y-x$

differentiating w.r.t time

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s$

7 0

3 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at

Soloha48 [4]

Answer:

v = 11.0 m/s at 198.6° (18.6° south of west)

ΔKE = -145 kJ

Explanation:

I assume you want to find the final velocity and the change in kinetic energy.

Take east to be +x and north to be +y.

Momentum is conserved in the x direction:

(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ

vₓ = -10.4 m/s

Momentum is conserved in the y direction:

(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ

vᵧ = -3.50 m/s

The magnitude of the final velocity is:

v² = (-10.4 m/s)² + (-3.50 m/s)²

v = 11.0 m/s

The direction of the final velocity is:

θ = atan(-3.50 m/s / -10.4 m/s)

θ = 198.6°

The initial kinetic energy is:

KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²

KE₀ = 253,275 J

The final kinetic energy is:

KE = ½ (1800 kg) (11.0 m/s)²

KE = 108,682 J

The change in kinetic energy is:

ΔKE = 108,682 J − 253,275 J

ΔKE ≈ -145,000 J

7 0

3 years ago

Name some applications of white surfaces.

timofeeve [1]

White marble is a white surface

7 0

3 years ago

Read 2 more answers

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and

snow_lady [41]

Answer:

Explanation:

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be $v_{P/A} = v( cos \theta \ i + sin \theta \ j)$

where:

$v_{P/A}$ = velocity of the plane in regard to the air

v = velocity

θ = angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

$v_{P} = v_o( cos \phi \ i + sin \phi \ j)$

where;

$v_p$ = velocity of the airplane

$v_o$ = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing $v_o$ = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

$v_p = v_A + v_{P/A}$

$v_A= v_P -v_{P/A}$ --- (1)

$v_A=$ ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

$v_A=$ (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

$v_A= (-39.24 km/h)i + (13.44 km/h) j$

$|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}$

$v_A$ = 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ = $tan ^{-1} (2.9)$

θ = 70.97°

The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.

5 0

3 years ago