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3 years ago

An object's acceleration is given by a(t)=a(t)=60t m/s260t m/s2 . if it begins at rest, how far has it gone after 10 seconds?

1 answer:

6 0

The object's acceleration is given by the expression: a(t) = 60t m/s^2. The distance it traveled after 10 seconds after starting from rest can be obtained by integrating the expression, a(t), twice in order to get the expression for the distance as a function of time.

The first integral of a(t) is the velocity, v(t), of the object which is then equal to: v(t) = 30t^2. The integral of v(t) is then distance, d(t), of the object which is then equal to: d(t) = 10t^3. With the distance equation available, we simply plug in t = 10 seconds into the equation. It is then determined that the object has traveled 10,000 m or 10 km after 10 seconds starting from rest.

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A revolutionary war cannon, with a mass of 2240 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 131 m/s aft

My name is Ann [436]

Answer:

v' = -0.0906 m/s

Explanation:

given,

mass of cannon, M = 2240 Kg

mass of the ball, m = 15.5 Kg

speed of cannon ball, v = 131 m/s

speed of he cannon = ?

initial speed of cannon and the cannon ball is equal to 0 m/s

using conservation of energy

(M+m)V = M v' + m v

(M+m) x 0= 22400 v' + 15.5 x 131

22400 v' = -2030.5

v' = -0.0906 m/s

negative sign represent the canon will move in opposite direction of the ball.

hence, speed of cannon is equal to 0.0906 m/s

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2 years ago

Which exoplanet has the most Earth-like orbit?

Digiron [165]

Answer: Kepler-452b

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4 0

2 years ago

If two oscillators connected to speakers create frequencies of 594 hz and 600 hz, what is the beat frequency when you hear both

Annette [7]

When we have two sound waves with frequencies f1 and f2 that interfere, the beat frequency is equal to the absolute value of the difference of the two frequencies, so in our problem:
$f_{beat}=f_2-f_1 = 600 Hz-594 Hz = 6Hz$

3 0

2 years ago

What is the speed of a 48-kilogram dog running across<br>a lawn with 216 joules of kinetic energy?

pishuonlain [190]

Answer:

3m/s

Explanation:

K.E= (1/2)mv^2

216j= (1/2)48kg • v^2

216J=24kg•v^2

v^2 = (216J)/(24kg)

v^2= 9m^2/s^2

/sqrt{v^2} = /sqrt{9m^2/s^2}

V =3m/s

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3 years ago

3. why is the sum of the maximum voltages across each element in a series r l c circuit usually greater than the maximum applied

Llana [10]

The sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

<h3>What is the Kichoff's loop rule?</h3>

Kirchhoff's loop rule states that the algebraic sum of potential differences, as well as the voltage supplied by the voltage sources and resistances, in any loop must be equal to zero.

In a series RLCcircuit, the voltages are not added by scalar addition but by vector addition.

Kirchhoff's loop rule is not violated since the voltages across different elements in the circuit are not at their maximum values.

Therefore, the sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

Learn more about Kichoff's loop rule at: https://brainly.in/question/35360816

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8 0

2 years ago