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Svetach [21]
2 years ago
8

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 12m/s, how far from the base of the dorm should

Allison stand to catch the ball? Assume the vertical distance between where Emily releases the ball and Allison catches it is 8.0m. Express your answer with the appropriate units.
Physics
1 answer:
Hitman42 [59]2 years ago
6 0

Answer:

3.6 m

Explanation:

let x = horizontal distance between emily and allison should be for allison to catch the ball

Find horizontal speed of the ball

vx = 12 sin 30 = 12 x 0.5 = 6 m/s

To find time taken, we will use vertical values of the ball motion

Initial velocity in vertical direction

u = 12 cos 30 = 10.392 m/s

let a = g = 9.8m/s2

Use equation of motion

s = ut +1/2at^2

s = vertical distance = 8

8 = (10.392)t + (1/2)(9.8)t^2

8 = (10.392)t + (4.9)t^2

4.9t^2 + 10.392t - 8 = 0

Using formula of quadratic or calculator, we'll find

t = 0.6 and t = -2.72

We pick t=0.6s since it's not logical time in negative

Assuming no air resistance or external forces, the ball will move 6m/s horizontally. Hence using the formula of speed

speed vx = distance x / time

x = (vx)(t)

  = 6 x 0.6

  = 3.6 m

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lilavasa [31]

Answer:

9.6x10¹³watt

Explanation:

See attached file

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3 years ago
Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t
Aliun [14]

kinetic energy is given as

KE = (0.5) m v²

given that : v = speed of the bottle in each case =  4 m/s

when m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

6 0
2 years ago
Read 3 more answers
Two weights are connected by a very light cord that passes over an 80.0Nfrictionless pulley of radius 0.300m. The pulley is a so
Citrus2011 [14]

Answer:

The force does the ceiling exert on the hook is 269.59 N

Explanation:

Applying the second Newton law:

F = m*a

From the attached diagram, the net force in object 1 is:

m_{1} a=T_{1} -W_{1}

In object 2:

m_{2} a=W_{2} -T_{2}

Adding the two equations:

m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2}   -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a  (eq. 1)

The torque:

\tau =I\alpha

Where

I = moment of inertia

α = angular acceleration

If the linear acceleration is

a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}

Torque due the tension is equal:

\tau =r(T_{2} -T_{1} )

Substituting torque, mass, in equation 1, the expression respect the acceleration is:

a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }

Where

W₁ = 75 N

W₂ = 125 N

W = 80 N

a=\frac{9.8*(125-75)}{75+125+\frac{80}{2} } =2.04m/s^{2}

The net force is:

F_{n} =F-W-T_{1} -T_{2}\\0=F-W-W_{1} (\frac{a}{g} +1)-W_{2} (1-\frac{a}{g})\\F=W+W_{1} +W_{2} +\frac{a}{g} (W_{1} -W_{2} )\\F=80+75+125+\frac{2.04}{9.8} (75-125)\\F=269.59N

4 0
2 years ago
Elyse is explaining to a friend that she will be conducting a scientific investigation. Her question is "How many leatherback se
trapecia [35]

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A one-dimensional conservative force is given by the function F(x) = (2.00 N/m) x +(1.00 N/m3) x3. What is the change in potenti
topjm [15]

Answer:

Explanation:

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P.E is given as

P.E=∫F(x)dx

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P.E= 2x²/2 + x⁴/4 from x=1 to x=2

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P.E=(2²+2⁴/4)-(1² +1⁴/4)

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Since it is moving from lower position to upper position, then P.E should be negative

P.E = - 6.75J

8 0
2 years ago
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