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natima [27]
3 years ago
8

Physics Help Please!!!

Physics
2 answers:
kolezko [41]3 years ago
8 0
v = 13 \frac{m}{s}, \theta = 54^\circ \\  \\ v_{xi} = v*cos \theta \\ v_{yi} = v*sin\theta \\  \\ a_x = 0, a_Y = -10 \frac{m}{s^2}  \\  \\ v_{yf} = v_{yi} + a_yt = -v_{yi} \\ t =  \frac{2v_{yi}}{a_y}

liq [111]3 years ago
3 0
The answer is the third graph
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A lamp can work on 50V mains taking 2 amps. What value of resistance must be connected in series with it, so that it can be oper
wolverine [178]
The resistance of the lamp is apparently  50V/2A  =  25 ohms.

When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.

In order for 200V to cause 2A of current, the total resistance
must be      200V/2A = 100 ohms.

The lamp provides 25 ohms, so we want to add another 75 ohms 
in series with the lamp.  Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.

The power delivered by the 200V mains is (200V) x (2A) = 400 watts.

The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.

The extra resistor dissipates  ( I² · R) = (2² · 75 ohms) = 300 watts.

Together, they add up to the 400 watts delivered by the mains.

CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it. 
It has to be a special 'power resistor'. 
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms.  Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
3 0
3 years ago
A 55.0-kg skydiver drew falls for a period of time before opening his parachute. what is his kinetic energy when he reaches a ve
Jlenok [28]
Mass (m)=55kg

acceleration (a)=9.81 m/s^2, this is the acceleration due to gravity.

initial velocity=0m/s. The skydiver doesn’t start with any speed because she is on the plane or helicopter.

final velocity=16m/s This is the velocity (speed) the skydiver reaches

The equation we use is KE=.5mv^2
Kinetic energy=.5 mass x velocity^2

KE=.5(55kg)(16m/s)^2
KE=.5(55kg)(256m/s)
KE=.5(14080J)

J=Joules

KE=7040J

Kinetic energy is 7040 Joules (J)

Hope this helps
3 0
4 years ago
Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
andreev551 [17]

Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

3 0
3 years ago
A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners
galina1969 [7]

Answer:

Explanation:

As the source is situated on x - axis , it must be situated in between the two listeners .

So the x coordinate of source is

(-7 + 3 )/2

= - 2 m

The equation of the wave- front will be that o a circle having centre at (-2,0)

and radius = distance between -2 and 3 , that is 5 m

equation of circle

=( x+2 )² + y² = 25

It cuts y axis when x = 0

Putting x = 0

4 + y² = 25

y² = 21

y = + √21 , or - √21

7 0
3 years ago
Vapor pressure is related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the _____. i
iVinArrow [24]

Answer

-Directly;  outside air pressure

Vapor pressure is directly related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the outside air pressure.

Explanation;

-As the temperature of a system increases, the average kinetic energy of the molecules increases in both the liquid and gas phases.

-A higher average kinetic energy facilitates the escape of molecules from the liquid phase into the gas phase. At the same time, the rate of return of gas phase molecules to the liquid also increases. A new equilibrium point is reached at a higher gaseous vapor pressure. The increase in vapor pressure with temperature is exponential.


6 0
3 years ago
Read 2 more answers
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