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xxTIMURxx [149]
3 years ago
10

A pilot claims to have seen a UFO moving initially at a speed of about 351 m/s in an easterly direction and then, in a time inte

rval of only 1.0 s, turning 48° south of east and moving at 351 m/s. Compute the UFO's average acceleration during the turn. Find Magnitude and direction
Physics
1 answer:
serious [3.7K]3 years ago
3 0

Answer:

Explanation:

This problem can be solved easily if we represent velocity in the form of vector.

The velocity of 351 was towards easterly direction so

V₁ = 351 i

The velocity of 351 was towards south west making - 48° with east or + ve x direction.

V₂ = 351 Cos 48 i - 351 sin 48 j

V₂ = 234.86 i - 260.84 j

Change in velocity

= V₂ - V₁ = 234.86 i - 260.84 j - 351 i

= -116.14 i - 260.84 j

acceleration

= change in velocity / time

(-116.14 i - 260.84 j )/ 1

= -116.14 i - 260.84 j

magnitude = 285.53 ms⁻²

Direction

Tan θ = 260.84 / 116.14 = 2.246

θ = 66 degree south of west .

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The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;
Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0
2 years ago
Which statement describes a surface wave?
nexus9112 [7]
The answer would be c because a surface wave travels between two different materials, like air and water.
8 0
2 years ago
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Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th
Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

\theta=30^{\circ}

So, the ball's horizontal velocity is

v_x = u cos \theta = (15)(cos 30)=13.0 m/s

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

y=u_yt+\frac{1}{2}at^2

where

u_y = u sin \theta = (15)(sin 30) = 7.5 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

d=v_x t =(13.0)(1.19)=15.5 m

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

d=98 m

while the horizontal velocity of the ball is

v_x = u cos \theta = (34)(cos 30)=29.4 m/s

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

y= h + u_y t + \frac{1}{2}at^2

where

h = 1.2 m is the initial height

u_y = u sin \theta = (34)(sin 30)=17.0 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

Substituting t = 3.33 s,

y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

u_y' = 7 m/s

So its position of the glove at time t' is

y'= h' + u_y' t' + \frac{1}{2}at'^2

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

t'' = t -t'

So we have two solutions:

t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he  falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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Answer:

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PLEASE HELP THIS IS DUE LIKE HELP ME PLEASE
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Answer:

At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,

The velocity will then increase to the midpoint of the motion.

m g h = 1/2 m v^2    where h is the vertical distance thru which the pendulum travels

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