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earnstyle [38]
3 years ago
12

2

Physics
1 answer:
Harman [31]3 years ago
6 0

Answer:

B

Explanation

Has the greatest deflection of the light ray from the initial ray direction.

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6. A football is kicked with an initial speed of 10.2 m/s at an angle of 40.00 above the horizontal. It lands on the ground 2.12
igor_vitrenko [27]

Answer:

7.81 m/s

Explanation:

Given,

initial speed, u = 10.2 m/s

angle of inclination, θ = 40°

time, t = 2.12 s

Horizontal component of the velocity:

u_x = u cos \theta

u_x = 10.2\times cos 40^0

u_x = 7.81 m/s

In projectile motion horizontal component of the velocity remain same at every point because there is no acceleration.

So, Velocity at the Pinnacle is equal to 7.81 m/s

7 0
3 years ago
Two people are standing on a 3.12-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a
Ann [662]

Answer:

The magnitude of displacement is 0.082m

Explanation:

While the ball is in motion,we have MV + mv= 0 ...eq1

Where M = combined mass of the platform and the two people.

V = velocity of the platform

m = mass of the ball

v = velocity of the ball

The distance that the platform moves is given by:

X = Vt ...eq2

Where t is the time that the ball is in the air.

The time the ball is in the air is given by:

L/(v-V) ...eq3

Where L is the length of the platform

The quantity(v-V) = velocity of the ball relative to the platform.

Combining eq2 and eq3

X = (V/(v - V))L

From eq1 , the ratios of the velocities is V/v = -m/M

X = (V/v)L / (1 - (V/v) = (-m/M) L /(1+ (m/M))

X = -mL/(M + m)

X = - (3.36kg × 3.12m) /( 119kg + 3.36kg)

X = - 10.48/ 122.36

X = -0.082m

The minus sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.

Therefore, the distance moved by the platform is the magnitude of this displacement 0.082m.

7 0
3 years ago
Read 2 more answers
In which arrangement of magnets will all the magnets attract
Elodia [21]

when they all are opisite

IDK I just guess

7 0
3 years ago
Read 2 more answers
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
IF YOU MOVE 50 METERS IN TO SECONDS,
Yanka [14]
You can use photo math for This
5 0
2 years ago
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