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Serggg [28]
4 years ago
5

A juggler is performing her act using several balls. She throws the balls up at an initial height of 4 feet, with a speed of 15

feet per second. If the juggler doesn't catch one of the balls, about how long does it take the ball to hit the floor?
Hint: Use H(t) = _16t2 + vt + s.

7.52 seconds
1.15 seconds
0.47 second
0.22 second

Physics
2 answers:
goblinko [34]4 years ago
6 0

Answer:

The ball will take 1.15 seconds to hit the floor.

Explanation:

It is given that,

A juggler is performing her act using several balls. The height as a function of time is given by :

h(t)=-16t^2+vt+s.............(1)

where

v is the speed of the ball

s is initial height

t is time taken

Putting h(t) = 0 and solving equation (1)

-16t^2+vt+s=0

-16t^2+15t+4=0

On solving above equation using graphing calculator and solving for t we get, t = 1.154 seconds

Hence, the correct option is (b) " 1.15 seconds ".

KatRina [158]4 years ago
4 0

<span>The correct answer between all the choices given is the second choice, which is 1.15 seconds. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
4 years ago
which of the following will not increase the strength of an electromagnet made by wrapping a wire around an iron nail
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what is the following?

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The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self pote
Alenkasestr [34]
It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
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8 0
3 years ago
A ball of mass 0.15 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. b)
Art [367]

Answer:

Explanation:

If v be the velocity just after the rebound

Kinetic energy will be converted into potential energy

1/2 m v² = mgh

v² = 2gh

v = √ 2gh

= √ 2 x 9.8 x .96

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4 0
3 years ago
A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeof
weeeeeb [17]

The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>
  • The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
  • V= final velocity, U= initial velocity, S= distance and a= acceleration
<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>
  • Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
  • So, 100²-0= 2×a×1800

=> 10000= 3600a

=> a= 10000/3600 = 2.8 m/s²

Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

#SPJ4

6 0
2 years ago
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