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3 years ago

For calcutating kinetic energy what units do the mass and speed have to be ( k.e=1/2mv^2 )

Physics

1 answer:

kiruha [24]3 years ago

7 0

Answer:

mass is in kg and velocity is in m/s for the answer to be in Joules

Explanation:

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2 years ago

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Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto

kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

The force of tugboat A, $F_A = 3000 N$ , acting in a certain direction
The force of tugboat B, $F_B = 5000 N$ , also acting in the same direction

Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

$F=F_A+F_B = 3000 + 5000 = 8000 N$

and the direction is the same as the direction of the two forces.

Learn more about forces:

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3 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo

yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is = $102^{\circ}-65.3^{\circ}=36.7^{\circ}$

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, $r^2$ representing the distance between the planes, we see that:

$r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295$

r = 741.6959 m

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3 years ago

Is this statement true or false?

Y_Kistochka [10]

The statement is true!

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3 years ago

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A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the subme

Elden [556K]

Answer:

The volume of the submerged part of her body is $0.0622m^{3}$

Explanation:

Let's define the buoyant force acting on a submerged object.

In a submerged object acts a buoyant force which can be calculated as :

$B=$ ρ.V.g

Where ''B'' is the buoyant force

Where ''ρ'' is the density of the fluid

Where ''V'' is the submerged volume of the object

Where ''g'' is the acceleration due to gravity

Because the girl is floating we can state that the weight of the girl is equal to the buoyant force.

We can write :

$W_{girl}=B$ (I)

Where ''W'' is weight

⇒ If we consider ρ = $1000\frac{kg}{m^{3}}$ (water density) and $g=9.81\frac{m}{s^{2}}$ and replacing this values in the equation (I) ⇒

$B=W_{girl}$

$B=610N$

ρ.V.g = 610N

$1000\frac{kg}{m^{3}}.V.(9.81\frac{m}{s^{2}})=610N$ (II)

The force unit ''N'' (Newton) is defined as

$N=kg.\frac{m}{s^{2}}$

Using this in the equation (II) :

$(9810\frac{N}{m^{3}}).V =610N$

$V=\frac{610N}{9810\frac{N}{m^{3}}}$

$V=0.0622m^{3}$

We find that the volume of the submerged part of her body is $0.0622m^{3}$

8 0

3 years ago

For calcutating kinetic energy what units do the mass and speed have to be ( k.e=1/2*m*v^2 )

For calcutating kinetic energy what units do the mass and speed have to be ( k.e=1/2mv^2 )