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3 years ago

A projectile is launched

1 answer:

4 0

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

initial horizontal velocity, vₓ = 16 m/s
initial vertical velocity, $v_y =0$
time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

$v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s$

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

Learn more here:brainly.com/question/20349275

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3 years ago

A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o

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Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: 

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Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.

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Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.

Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)

W = ∫ 1250π x dx
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4 years ago

a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on

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Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

$F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}$

Now, if the body is on the surface of the Earth, its weight (w) would be:

$F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}$

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

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