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Murrr4er [49]
3 years ago
15

Suppose a boat moves at 12.0m/s relative to the water. If the boat is in a river with the current directed east of 2.50m/s, what

is the boat's speed relative to the groiund when it is heading (a) east, with the current, and (b) west, against the current?
Physics
1 answer:
inessss [21]3 years ago
8 0

Answer:

(a) 14.5 m/s

(b) 9.5 m/s

Explanation:

Let the speed of the boat in the still water (i.e relative to ground) is v_0.

Given that the speed of the boat relative to water, v_r=12.0 m/s.

Speed of the water current of the river relative to the ground, u_0=2.50 m/s towards east.

(a) When the boat is heading toward the east (in the same direction of the current of the river)

The relative velocity, v_r, of the boat with respect to ground is

v_r=v_0-u_0

\Rightarrow 12=v_0-2.5

\Rightarrow v_0=12+2.5=14.5 m/s

Hence, the boat's speed relative to the ground when it is heading east, with the current, is 14.5 m/s.

(b) When the boat is heading toward the west (in the opposite direction of the current of the river)

The relative velocity, v_r, of the boat with respect to ground is

v_r=v_0-(-u_0)

\Rightarrow 12=v_0+2.5

\Rightarrow v_0=12-2.5=9.5 m/s

Hence, the boat's speed relative to the ground when it is heading west, against the current, is 9.5 m/s.

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Answer:

The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules

Explanation:

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3 years ago
You push a 85 kg shopping cart from rest with a net force of 250 n for 5 seconds,at which point it flies off a cliff that is 100
Vikki [24]

m = mass of the cart = 85 kg

F = net force on the cart = 250 N

a = acceleration of the cart

acceleration of the cart is given as

a = F/m

a = 250/85

a = 2.94 m/s²

t = time for which the force is applied = 5 sec

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart just before  it flies off the cliff = ?

using the equation

v = v₀ + a t

inserting the values

v = 0 + (2.94) (5)

v = 14.7 m/s

consider the motion of cart after it flies off the cliff in vertical direction :

v' = initial velocity in vertical direction = 0 m/s

a' = acceleration in vertical direction = g = acceleration due to gravity = 9.8 m/s²

t' = time taken for the cart to land = ?

Y' = vertical displacement of the cart = height of cliff = 100 m

using the kinematics equation

Y' = v' t' + (0.5) a' t'²

100 = (0) t' + (0.5) (9.8) t'²

t' = 4.52 sec


consider the motion of cart along the horizontal direction after it flies off the cliff

X = distance traveled from the base of cliff = ?

t' = time of travel = 4.52 sec

v = velocity along the horizontal direction = 14.7 m/s

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X = 14.7 x 4.52

X  = 66.4 m


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Dennis_Churaev [7]
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Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the
IRISSAK [1]

Answer:

a₁ = 0.63 m/s²  (East)

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F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

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⇒  a₁ = (60N / 95Kg) = 0.63 m/s²  (⇒)  East

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∑Fx = m*a   (⇒)

F = -m₂*a₂         ⇒      60 N = -51 Kg*a₂    

⇒  a₂ = (60N / 51Kg) = -1.18 m/s²  (West)

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