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3 years ago

8

Una pelota de béisbol es arrojada verticalmente hacia arriba desde la azotea de un edificio con una velocidad de 20m/s. Calcular

1 answer:

ExtremeBDS [4]3 years ago

5 0

buenos dias senor, en ingles por favor/per favore ???

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Imagine that you and a friend are discussing your health-related fitness levels. What assessments could you and your friend

Westkost [7]

Answer: In order to help improve your fitness levels you can swim, jog/run, body weight exercises, and a balanced diet. Explain why the greatest benefits to cardiorespiratory fitness come from sustained physical activities like running, walking and cycling.

Explanation:

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3 years ago

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How does friction affect the distance of an object?

stiv31 [10]

It slows the object down so it cannot move well and evetually the object cannot be pushed and farther

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3 years ago

Two forces act on a moving object that has a mass of 27 kg. One has a magnitude of 12 N and points due south, while the other ha

levacccp [35]

0.77 m/s2 directed 35° south of west

net force = (-17,-12)

net force = mass * acceleration

(-17,-12) = 27 * (x-acceleration,y-acceleration)

(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)

angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.

magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)

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2 years ago

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

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3 years ago

Fossil fuels store energy from the sun as<br><br> question 5

qwelly [4]

Answer:

Fossil fuels store energy from the sun as

Explanation:

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2 years ago