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KiRa [710]
1 year ago
11

A ball dropped from a window strikes the ground 2.71 seconds later. How high is the window above the ground?

Physics
1 answer:
koban [17]1 year ago
4 0

s=36.52m

Given ,

Time =2.71 second

We apply the formula

s=ut+12at^2

u=0,t=2.73  \\a=g \\So, the  \\height \\ s=36.52m

When a body is lying on a horizontal surface, the object is subjected to a downward force due to the action of gravity. These gravitational forces always aim to attract the item, confirming the force's nature as an attractive force. Consider an object of mass "M" resting on a horizontal surface; the force exerted to the object by gravity is equal to the weight of the object "W." The acceleration owing to gravity, indicated as "g," is the acceleration by which an object is drawn towards the ground. As a result, Newton's second law states that when the values in the above connection are substituted.

To learn more about Motion due to gravity from the given link:

brainly.com/question/14047086

#SPJ9

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An airplane flies at airspeed (relative to the air) of 280 km/h . The pilot wishes to fly due North (relative to the ground) but
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Answer:

the pilot should head the plane 7.547^{\circ} towarrds south- west

Solution:

The airspeed of the airplane, v = 280 km/h

The velocity of the wind, v' = 52 km/h South-west

Angle, \theta = 225^{\circ}

Now, measured angle in the clockwise direction from North:

sin225 = sin(\pi + 45) =  - sin 45^{\circ}

Now,

vsinx - v'sin45 = 0

280sinx = 52sin45

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x = 7.547^{\circ} south- west

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In fig. 30.11, suppose that and with switch open, switch is left closed until a constant current is established. then is closed
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In a particle accelerator a positron (C= +1.6 x 10-19) travels through a perpendicular magnet field with a magnitude of 3.1 x 10-2 T. At what speed must the positron travel in order for it to experience a force of 4.75 x 10-14 N? 28. An alpha particle (2 protons and 2 neutrons) experiences a downward force of 2.9 x 10-14 N while traveling in a magnetic field with a strength of 5.1 x 10-19 T pointing to the north. Find the speed of the particle and the direction it must be traveling in. 29. Find the length of a wire if it experiences a .63N force when it travels through a magnetic field with a strength of 0.85T whilst carrying 5.0 amps of current. 30. A coil with 462 turns of wire, a total resistance of 36Ω , and a cross-sectional area of 0.25 m2 is positioned with its plane perpendicular to the field of a powerful electromagnet. What average current is induced in the coil during the 0.37s that the magnetic field drops from 3.1 T to 0.0 T? 31. A step-up transformer has a potential difference across the primary of 28 V and a potential difference across the secondary of 3.0 × 104 V. There are 28 turns in the primary coil. How many turns are in the secondary? 32. A step-up transformer is used to create a potential difference of 1.6872 × 105 V across the secondary. The potentia
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3 years ago
Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_{1} +q_{2} =-q_{3} -----eqution 1

where,

q_{1} is the heat absorbed by the solid at 0⁰C

q_{2} is the heat absorbed by the liquid at 0⁰C

q_{3} the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_{2} = -q_{3}

<u>Step 4:</u> calculate the final temperature of the water

79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}

Substitute in the values; we will have,

79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})

79.13 kJ + 990.66J* (T_{f}-218}) = -1463J*(T_{f}-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_{f}-218}) = -1.463KJ*(T_{f}-100})

79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3

collect like terms,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0
3 years ago
50 POINTS PICTURE WITH QUESTION AND CHOICES BELOW
ad-work [718]
5 m/s sir I believe so
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