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Aleonysh [2.5K]
4 years ago
12

romeo threw a small bundle toward Juliet on a balcony 10.0 m above him with a velocity of 5.0 m/s. The bundle stopped rising aft

er 1.5 seconds. How high did the bundle travel? Was that high enough for her to catch it?
Physics
1 answer:
choli [55]4 years ago
4 0

I can tell that you've never read much Shakespeare.

If the bundle is pitched with a velocity of 5.0 m/s, it must follow
as the night follows the day that it cannot then continue to rise
for more than barely 1/2 second before it reacheth the zenith
of its arc, reverseth its course, and plungeth Earthward to its
ultimate demise.

But hey !  To thine own self be true, eh ?

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No force is necessary to keep a moving object moving (in a straight line at a constant speed).
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(marking brainliest) pease help asap! both of the questions are in the pdf, and please let me know which is question one and whi
REY [17]
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<em>1) The relationship in between the electrical energy carriesd by the transmission wires and the amount of the heat loss in it is due to the reason that when the electricity is flown through the wires there are some resistance found in these wires which creates a disturbance in the efficient flow of electricty.Also we know that current have an heating effect when it is in motion as due to if a large amount or magnitude of electricity is flown through the transmission wires it will carry a larger heat effected and also due to the resistance is provided by the wires and so the process of heat loss takes place.</em>

<em>2)It is important to minimize current in transmission wires due to minimize the heat loss and resistance on flowing electric current to make the system more efficient </em>

<em><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u></em><em> 3)Given Resistance = 250 ohms </em>

<em>Electric potential = 150 volts </em>

<em>so we know Power = </em>

<em>volt^2/Resistance = </em>

<em>=</em><em>(150^2/250)(ohms/volts)</em>

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<em>4)Heat energy (H) = Power(P)×Time(t)= (90×2)joules = 180 joul</em><em>e</em><em>s</em>

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4 0
3 years ago
A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re
stepan [7]
Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
 (20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
 (20 m/s)*sin(37°) = 12.036 m/s

The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by 
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy  = 12.036 - 9.8*0.2815
   = 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is 
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m  is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is 
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

6 0
3 years ago
If temperature increases, then pressure __________.
lilavasa [31]
If the temperature increases, then pressure increases too. (T<span>he molecules in the gas move faster, exerting a greater force. This </span>increases t<span>he </span>pressure<span>.)</span>
6 0
3 years ago
A sound source emits 20.0 W of acoustical power spread equally in all directions. The threshold of hearing is 1.0 × 10-12 W/m2.
ahrayia [7]

Answer:

0.0018 W/m²

Explanation:

Power and intensity are related as:

I=\frac{P}{4\pi r^2}

P=  20.0 W (given)

r = 30.0 m (given)

I=\frac{20.0}{4\pi(30.0)^2}=0.0018 W/m^2

Intensity in decibels:

I(dB)=10log\frac{I}{I_o}\\=10log\frac{0.0018}{10^{-12}}=92.5 dB

7 0
3 years ago
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