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Kitty [74]
3 years ago
6

A ball rolling across the floor at a velocity of 6.15 m/s [E] slows to rest in 1.00 minute. The

Physics
1 answer:
Vitek1552 [10]3 years ago
4 0

Answer: 0.102m/s^2

Explanation:

velocity of ball = 6.15 m/s

Time taken for ball to roll = 1.00 minute

Convert 1.00 minute to seconds

(Since 60 seconds = 1 minute, the time taken for ball to roll = 60 seconds

Acceleration of the ball = ? (Let unknown value be Z)

Since acceleration is the rate of change of velocity per unit time

i.e Acceleration = Velocity / Time

Z = (6.15 m/s / 60 seconds)

Z = 0.1025m/s^2

Round Z to 3 significant digits (Z = 0.102m/s^2)

Thus, the acceleration of the ball is 0.102 metres per second square

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A man can throw a ball a maximum horizontal distance of 56.1523 m. The acceleration of gravity is 9.8 m/s 2 . How far can he thr
IceJOKER [234]

Answer:

hmax= 28.2 m

Explanation:

Horizontal movement:

  • For the horizontal movement, as once released, nothing exerts any influence on the ball in the horizontal direction, it will keep moving at a constant speed, so, the equation for the horizontal displacement is as follows:

        x = v₀*t = 56.1523 m

Vertical movement:

  • For the vertical movement, once thrown upward, the ball is under the sole influence of gravity, g, which causes the ball to accelerate downward, and which magnitude is 9.8 m/s².
  • At any time, applying the definition of acceleration, we can find the value of the velocity, as follows:

        vf = vo-g*t

  • When the ball reaches to the maximum height, the ball will come momentarily at rest, so vf =0.
  • At this point, we can find the time at which  the ball reached to its highest point, as follows:

        t = \frac{vo}{g}

  • In the same way, we can find the maximum height reached by the ball, just replacing this value of time in the equation for the displacement, in the vertical direction, as follows:

        h = vo*t -\frac{1}{2} *g*t^{2}

        hmax = vo*(\frac{vo}{g}) -\frac{1}{2} *g*(\frac{vo}{g} )^{2} =\frac{vo^{2}}{2*g}

  • Now, returning to the horizontal movement, as the time must be the same for both movements, we can replace the value for time we have just found, in the equation for the horizontal displacement:

       x = v0x * t = v0x* \frac{voy}{g}

  • But as we know that vox = voy, we can rewrite the equation above as follows:

       x = v0* t = v0* \frac{vo}{g} =\frac{vo^{2} }{g} = 56.1523 m

  • We can solve for v₀, as follows:

        vo = \sqrt{56.1523 m *9.8 m/s2} =23.5 m/s

  • Now, we can replace this value in the expression for hmax, as follows:
  • hmax = \frac{vo^{2}}{2*g} =\frac{(23.5m/s)^{2}}{2*9.8 m/s2} =  28.2 m
4 0
3 years ago
Planet B has a tilt of 45 degrees. What seasonal changes would be expected on this planet?
12345 [234]
... Extreme temperature changes between seasons.

... Extreme changes in length of daylight and darkness.

... Larger areas around both poles where daylight/darkness
can last more than a whole day. 
Altogether, these areas cover half of the planet. 
5 0
3 years ago
An object with 274 J of GPE is 140cm above the ground. What is its mass?
Amanda [17]
E=274J
h=140cm=1,4m
g≈9,8m/s²

m=?

E=mgh
m=E/gh=274J/9,8m/s²*1,4m≈20kg



"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."



Regards M.Y.
3 0
3 years ago
Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco
11Alexandr11 [23.1K]

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

5 0
3 years ago
A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg · m 2 is rotating freely with an angular speed of 1
lidiya [134]

Answer:

angular speed = 0.4 rad/s

Explanation:

given data

radius = 5 m

moment of inertia = 2000 kg-m²

angular speed = 1.0 rad/s

mass = 60 kg

to find out

angular speed

solution

Rotational momentum of merry-go-round = I?

we get here momentum that is express as  

momentum = 2000 × 1

momentum = 2000 kg-m²/s

and  

Inertia of people will be here as

Inertia of people = mr² = 60 × 5²

Inertia of people = 1500 kg-m²

so Inertia of people for two people  

1500 × 2 = 3000

and

now conserving angular momentum(ω)

moment of inertia × angular speed = ( momentum + Inertia of people ) angular momentum  

2000 ×  1 = (2000 + 3000 ) ω

solve we get now  

ω = 0.4 rad/s  

5 0
3 years ago
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