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Masteriza [31]
3 years ago
13

Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

ada’s velocity to be 8 m/s ˆ I + 3 m/s ˆ j.
What is Jada’s speed with respect to the water?
a. 13.3 m/s
b. 11.3 m/s
c. 8.54 m/s
d. 8.25 m/s
e. 4.24 m/s
f. None of these
Physics
1 answer:
olchik [2.2K]3 years ago
5 0

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= \sqrt{(8)^2+(-2)^2}

|v|= \sqrt{68} = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

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4 years ago
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
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Answer:

d) g/2

Explanation:

We need to use one of Newton's equations of motion to find the position of the stone at any time t.

x(t) = x₀(t) + ut - ¹/₂at²

Where

x₀(t) = initial position of the stone.

x(t) - x₀(t) = distance traveled by the stone at any time.

u = initial velocity of the stone

a = acceleration of the stone

t = time taken

On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.

=> 0 = x₀(t)

=> x₀(t) = 0

On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.

=> 0 = 0 + uT - ¹/₂gT² (a = g)

=> uT = ¹/₂gT²

=> g = 2u/T

On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.

=> 0 = 0 + u(2T) - ¹/₂a(2T)²

=> 2uT = 2aT²

=> a = u/T

By comparing we see that a = g/2.

5 0
3 years ago
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kvv77 [185]

Answer:

The Autocratic system of governance

Explanation:

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