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3 years ago

An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object? (Formula: )

Physics

1 answer:

andrezito [222]3 years ago

6 0

Answer:

1/2mv^2 a(7.5 b(15 kg c(60 kg d(120 kg. nevermind i found the answer its (15 kg) because to solve for m its m= K2/v squared.

Explanation:

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A truck traveling with an initial velocity of 22m/s comes to a stop in 17.32 secs. What is the acceleration of the truck?

Tomtit [17]

Explanation:

Given:

v₀ = 22 m/s

v = 0 m/s

t = 17.32 s

Find: a

v = at + v₀

(0 m/s) = a (17.32 s) + (22 m/s)

a = -1.270 m/s²

Round as needed.

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2 years ago

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Which statement about projectile motion is true?

Svetllana [295]

C. The range of a projectile increases with an increase in the angle of launch.

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3 years ago

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A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along

Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

$F(x)=(3\ N/m^2)x^2+(6\ N/m)x$

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

$W=\int\limits {F{\cdot} dx}$

$W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J$

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

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3 years ago

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A baseball is thrown at a 28° angle and an initial velocity of 70 m/s. Assume no air resistance. What is the vertical component

icang [17]

Answer:

answer is 61.8 m/s; 32.9

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2 years ago

Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains

tester [92]

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v = (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

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3 years ago