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Julli [10]
3 years ago
6

A ball is thrown horizontally from a cliff and hits the ground 4 seconds later 40 meters from the base of the cliff. How high wa

s the cliff (rounded to the nearest tenth of a meter)?
Physics
2 answers:
DochEvi [55]3 years ago
8 0
<span>A ball is thrown horizontally from a cliff at a speed of 15 m/s and strikes the ground 45 meters from the base of the cliff. How high was the cliff (rounded to the nearest meter)</span>
sashaice [31]3 years ago
7 0

Answer:

h= 78.4 m

Explanation:

The ball moves with a uniformly accelerated movement in the vertical direction (y), we apply the following formulas:

vfy= v₀y+g*t Formula (1)

vfy²=v₀y²+2*g*h Formula (2)

h: hight in meters (m)    

t : time in seconds (s)

v₀y: initial speed in y  (m/s)  

vfy: final speed in y ( m/s   )

g: accelerationdue to gravity  (m/s² )

Known Data

v₀y= 0

t= 4 s

g= 9,8 m/s²

We apply the formula (1) to calculate vfy

vfy= v₀y+g*t

vfy= 0+ (9,8)*(4)

vfy= 39.2 m/s

We apply the formula (2) to calculate h

vfy²=v₀y²+2*g*h

(39.2)²=0+2*9.8*h

h = (39.2)²  /  (2*9.8)

h= 78.4 m

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Do 112m /29s which it will be 3.862 which if you round it, it will be 3.86 m/s
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In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
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Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

Speed of cart   V=5.00m/s

Mass of package   M_p=15.0kg

Speed of package at end of chute V_c=3.00m/s

Angle of inclination   \angle =37

Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

  \frac{1}{2}u^2+gh=\frac{1}{2}v_p^2

  v_p=\sqrt{2(\frac{1}{2}u^2+gh)}

  v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}

  v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}

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4 0
2 years ago
Why do planets rings surround them and not the sun
Shalnov [3]

Well, the rings surrounding a planet are made out of rock. A ring surrounding the sun would be impossible since the sun can reach more than 27 million degrees Fahrenheit (15 million degrees Celsius.)

Hope this helped.

7 0
3 years ago
if a Firebird travels at a velocity of 0 to 60 mph in four seconds traveling east what was the acceleration of the Firebird
Tresset [83]

Answer:

6.7 m/s^2

Explanation:

The formula of acceleration is:

\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{v_2 - v_1}{t_2-t_1}}

where \displaystyle{\vec{a}} is acceleration, \displaystyle{\vec{v}} is velocity and \displaystyle{t} is time. \displaystyle{v_2} means final velocity. \displaystyle{v_1} means initial velocity, \displaystyle{t_2} means final time and \displaystyle{t_1} means initial time.

We are given that the Firebird travels at velocity of 0 to 60 mph in four seconds. Therefore:

  • Our initial velocity starts at 0 mph.
  • Our final velocity is at 60 mph.
  • Our initial time is 0 second.
  • Our final time is 4 seconds.

Since it travels to the east then our vector will be positive. However, acceleration has to be in m/s^2 unit (Sl unit) so we'll have to convert from mph (miles per hours) to m/s (meters per second) first.

We know that:

  • A mile equals to 1609.344 meters.
  • An hour equals to 60 minutes which a minute equals to 60 seconds. So 60 minutes will equal to 3600 seconds.

Now we divide 1609.344 by 3600 to find a unit rate of m/s:

\displaystyle{\dfrac{1609.344}{3600} \ \, \sf{m/s}}\\\\\displaystyle{= 0.44704 \ \, \sf{m/s}}

Now multiply 0.44704 m/s by 0 and 60 to get velocity in m/s unit:

  • Initial velocity = 0 m/s
  • Final velocity = 60 * 0.44704 = 26.82 m/s

Time is already in second so no need for conversion. Substitute known information in the formula:

\displaystyle{\vec{a} = \dfrac{26.82-0}{4-0}}\\\\\displaystyle{\vec{a} = \dfrac{26.82}{4}}\\\\\displaystyle{\vec{a} = 6.7 \ \, \sf{m/s^2}}

Therefore, the Firebird will accelerate at the rate of 6.7 m/s^2.

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_____ measures an objects change in position per unit time.
Anettt [7]
Speed (ex: meters/second, miles/hour)
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