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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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oee [108]
The correct order is (in decreasing order of gravity strength)
Jupiter - Neptune - Venus - Mars

In fact, Wayne's weight on each planet is given by
W=mg
where m is Wayne's mass, which is a constant value, and g is the gravity strength at the surface of the planet. Therefore, the Wayne's weight W on each planet is directly proportional to the gravity strength of that planet: so the planet with the strongest gravity is the one where Wayne's weight is the greatest (Jupiter, 333 pounds), followed by Neptune (159),  Venus (128) and Mars (53).
8 0
2 years ago
Briana swings a ball on the end of a rope in a circle. The rope is 1.5 m long. The ball completes a full circle every 2.2 s. Wha
schepotkina [342]
The radius of the circular path is 1.5 m.

The circumference is then
1.5\ m*2\pi=3\pi\ m

The ball moves 3π m every 2.2 s, so the speed is
\frac{3\pi\ m}{2.2\ s}\approx 4.3\ m/s
9 0
2 years ago
Read 2 more answers
The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan
PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

B=\frac{\mu _o I}{2\pi r}

where B= magnetic field

           \mu _o= permeability of free space

           I= current in the long wire and

           r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

   

7 0
2 years ago
Read 2 more answers
You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac
Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1  + \frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1  + \frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}

Stop sight distance= 81.88 m

5 0
2 years ago
John is going to use a rope to pull his sister Laura across the ground in a sled through the snow. He is pulling horizontally wi
bulgar [2K]

Answer:

The mass of Laura and the sled combined is 887.5 kg

Explanation:

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F_T = F_L+F_S

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From Newton's second law of motion, "the rate of change of momentum is directly proportional to the applied force.

F_T = \frac{(M_L+M_S)V}{t}

where;

M_L is mass of Laura and

M_S is mass of sled

Mass of Laura and the sled combined is calculated as follows;

(M_L+M_S) = \frac{F_T*t}{V}

given

V = Δv = 4-0 = 4m/s

t = 5 s

(M_L+M_S) = \frac{710*5}{4}\\\\(M_L+M_S) =  887.5 kg

Therefore, the mass of Laura and the sled combined is 887.5 kg

4 0
2 years ago
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