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const2013 [10]
3 years ago
7

A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb

oat? b) If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? c) What distance downstream does the boat reach the opposite shore?
Physics
1 answer:
Nostrana [21]3 years ago
4 0

Explanation:

It is given that,

Velocity in East, v_1=4\ m/s

Velocity in North, v_2=3\ m/s

(a) The resultant velocity is given by :

v=\sqrt{4^2+3^2}=5\ m/s

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

t=\dfrac{d}{v}

t=\dfrac{80\ m}{5\ m/s}

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

x=v_2\times t

x=3\ m/s\times 16\ s

x = 48 meters

Hence, this is the required solution.

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If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?
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Hi there!

\large\boxed{\text{B) 20 meters}}

We know that:

E_T = U + K

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

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Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

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3 years ago
A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x
mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

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a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

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x = 12.5 - 0

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