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zalisa [80]
3 years ago
13

A 10-kilogram box is at static equilibrium, and the downward pull of gravity acting on the box is 98 newtons. What is the minimu

m force that would be required to just pick up the box?
Physics
2 answers:
polet [3.4K]3 years ago
5 0

Answer:

slightly greater than 98 newtons in the upward direction

Explanation:

this is a plato answer

Nikolay [14]3 years ago
4 0
The box is at equilibrium, so the net force on the box is zero (the force of gravity on the box is equal to the force exerted up on the box by the surface on which it rests.)
To pick up the box, our upward force must be greater than the force of gravity on the box (the weight). So, we must lift up the box with a force greater than 98 newtons. :)
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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
Nostrana [21]

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.

4 0
3 years ago
Explain two ways in which humans are related<br>to whales.​
Maksim231197 [3]

Answer:

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4 0
2 years ago
Describe the two main ways that energy travels through the Sun.
ValentinkaMS [17]

Answer:

The energy are traveled through the sun in two ways that are:

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As, the energy are first traveled outward from the core of the sun and then it get entered in the zone of the radiation and this zone is known as radiative zone in the surface of the sun.In the radiation process some amount of the hydrogen particle are combined and then releases the energy. And the convection is the process of conversion of one form of energy to another form.

7 0
3 years ago
In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of t
a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot,  at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot  = Pgas

        = 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵  =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

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ΔP = ρgh

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h = 65.85 meters

7 0
3 years ago
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