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8090 [49]
2 years ago
12

1. A student adds water and sugar to a jar and seals the jar so that nothing can get in or out. The

Physics
1 answer:
Tcecarenko [31]2 years ago
3 0

Answer:

The mass will stay the same throughout time

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The graph shows two runners participating in a race.
Olin [163]

Answer:

'Daniela had a 5-meter head start, and Leonard caught up to her at 25 meters.'

Explanation:

hope that helps :)

6 0
2 years ago
Read 2 more answers
2. A 50 kg diver is standing on the edge of a 15 m high cliff. What is his potential energy?
Lera25 [3.4K]

Answer:

3675 J

Explanation:

Gravitational Potential Energy = \frac{1}{2} × mass × g × height

( g is the gravitation field strength )

Mass = 50 kg

G = 9.8 N/kg ( this is always the same )

Height = 15 m

Gravitational Potential Energy = \frac{1}{2} × 50 ×9.8 × 15

= 3675 J

3 0
3 years ago
Two identical positive charges are placed near each other. At the point halfway between the two chargesTwo identical positive ch
Nitella [24]

Answer:The electric field is zero and the potential is positive.

Explanation:

Two identical positive charges are separated by a certain distance and midway between charges two identical positive charges are placed near each other.

So the Electric field at midway is zero because the electric field due to both charges add up to give zero electric field.(because they point in opposite direction)

Potential is scalar quantity and charges are positive so they add up to give potential.

7 0
3 years ago
g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
3 years ago
Answer question 83 and 84
Ne4ueva [31]
83. Gravity is activity as a force on the cart.

84. The box has gathered kinetic energy and momentum and will continue going forward.
6 0
3 years ago
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