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4 years ago

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

Physics

1 answer:

aksik [14]4 years ago

8 0

Answer: 2200J

Explanation:

M = 44kg

V = 10m/s

K.E =?

K.E = 1/2MV2 = 1/2 x 44 x (10)^2

K.E = 22 x 100

K.E = 2200J

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The stars, Rigel and Betelgeuse, are both found in the constellation, Orion. Rigel is a blue supergiant, and Betelgeuse is a red

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3 years ago

calculate the value of net force acting on given body?5N acting towards right & 15N force acting towards left on the brick.(

lys-0071 [83]

Answer:

The net force acting on the body is 10N directed to the left.

Explanation:

Magnitude of force to the right = 5N

Magnitude of force to the left = 15N

Net force acting on the object and in what direction;

Solution:

It is the vector sum of all forces acting on a body. This net force is the single force that will replace the forces acting on a body;

For the problem;

Net force = Force to the left + Force to the right

Let us take left to be negative and right to be positive;

Force to the left = -15N

Net force = -15N + 5N = -10N

The net force acting on the body is 10N directed to the left.

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3 years ago

The note created by a flute will increase the speed of sound increases. When a marching band goes outside on a cold day, what wo

alexgriva [62]

A).

It would decrease because the speed of sound and temperature are proportional.

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4 years ago

Read 2 more answers

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

4 years ago