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sveticcg [70]
3 years ago
8

A boat is traveling upstream in the positive direction of the x axis at 10 km/h with respect to the water of a river. The water

is flowing at 7 km/h with respect to the ground. (a) What is the velocity of the boat with respect to the ground?
Physics
1 answer:
sasho [114]3 years ago
4 0

Answer:

The velocity of the boat with respect to the ground is 3 km/h

Explanation:

The speed of an object is different depending on the reference system you use. This is called relative speed.

A boat travels upstream, this means that it moves in the opposite direction to the river current.

A boat travels upstream, this means that it moves in the opposite direction to the river current. Then, if the boat moves in the positive direction of the x axis at 10 km / h with respect to the water of a river, the water flows in the negative direction of the x axis at 7 km / h with respect to the ground.

This causes the speed of the boat relative to the ground to be calculated as follows:  

<em>VbG = Vbw - VwG </em>

where VbG is the speed of the boat relative to the ground, Vbw is the speed of the boat relative to the water of the river and VwG is the speed of the water relative to the ground.

So: VbG=10 km/h – 7 km/h

<u><em>VbG= 3 km/h </em></u>

The direction of this velocity is in the positive x-direction.

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REY [17]

Answer:

Explanation:

El impulso aplicado a la pelota produce una variación en su momento lineal.

J = m (V -Vo)

Conviene elegir positivo el sentido de la velocidad final.

J = 0,100 kg [40 - (- 20)] m/s = 6 kg m/s

Saludos Herminio

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Explanation:

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8 0
3 years ago
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Evgesh-ka [11]

Answer:

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8 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

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5 0
2 years ago
The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point
makvit [3.9K]
To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.
5 0
3 years ago
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