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Ludmilka [50]
3 years ago
10

PLEASE HELP THE question on picture

Physics
1 answer:
Gnoma [55]3 years ago
7 0

Vertical:

(20 m/s) sin(25º) ≈ 8.45 m/s

Horizontal:

(20 m/s) cos(25º) ≈ 18.1 m/s

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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

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While fishing on a lake, a fisherman notices ripples 0.32 m apart, and they are
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the main properties of the main wave propertioes include wavelength amplitude, cruest an trough

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Why quantity refractive index dooesn't have unit
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Refractive Index is a ratio of two similar physical quantity which is dimension less
refractive index = sin I / sin r

therefore it doesn't have a unit.
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2 years ago
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An airplane flies with a constant speed of 840 km/hr. How far will go in 270 min
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3780 km.
..................
.....
........












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3 years ago
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At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v
Anna007 [38]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. For this purpose we will define the speed as the distance traveled in a given period of time. Here the distance is equivalent to the orbit traveled around the earth, that is, a circle. Approaching the height of the aircraft with the radius of the earth, we will have the following data,

R= 6370*10^3 m

v = 219m/s

a = 17m/s^2

The circumference of the earth would be

\phi = 2\pi R

Velocity is defined as,

v = \frac{x}{t}

t = \frac{x}{v}

Herex = \phi, then

t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{219}

t = 1.82*10^5s

Therefore will take 1.82*10^5 s or 506 hours, 19 minutes, 17 seconds

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2 years ago
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