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sertanlavr [38]
3 years ago
7

Take a look at the weather map. The front seen there causes short periods of storms and heavy rains. What type of front is this?

Physics
2 answers:
erastova [34]3 years ago
5 0
I am pretty sure it is a cold front.
Naily [24]3 years ago
5 0
A-cold front  hopes this helps

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At Hoover Dam, the distance the water effectively falls before encountering the electric generators depends on the water levels
pashok25 [27]

Answer:

Explanation:

recall that power is energy carried out or work done per time

P=W/t

P=2*10^6*35

t=6*60=420S

W=Energy

E=2*10^6*35*360S

E=25200000000

Energy stored by water from rest is called potential energy. Since the water is falling from a height , we calculate potential energy as thus

E=M*g*h

Assume that the water intakes are effectively 175 m above the electric generators. How much water must pass through the generators to power 2 million 35-W Las Vegas light bulbs for 6.0 minutes?

M=mass of water

g=acceleration due to gravity 9.81m/s^2

h=height ,175m

25200000000=M*9.81*175

M=\frac{25200000000}{175*9.81}

M=1716.75kg

3 0
3 years ago
What are the invisible forces of nature
Tems11 [23]

Answer:

Gravity, Friction, Air resistance, magnetism, static electricity

Explanation:

3 0
4 years ago
Read 2 more answers
An object is moving at a constant velocity of 10 m/s for 5 seconds. What is
german

initial velocity (u)=0m/s

final velocity (v)=10m/s

time( t)=5s

acceleration (a)=v-u÷t

acceleration (a)=10-0÷5

acceleration (a)=10÷5

acceleration (a)=2

therefore acceleration (a)=2m/s

3 0
3 years ago
A bullet is fired straight up from a gun with a
melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

V=V_{o}+gt (1)

V^{2}=V_{o}^{2}+2gy (2)

Where:

V is the final velocity of the bullet

V_{o}=152 m/s is the initial velocity of the bullet

g=-9.8 m/s^{2} is the acceleration due gravity, always directed downwards

t=6.9 s is the time

y is the vertical position of the bullet at t=6.9 s

Let's begin by finding V from (1):

V=152 m/s-9.8 m/s^{2}(6.9 s) (3)

V=84.38 m/s (4)

Now we have to substitute (4) in (2):

(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y (5)

Isolating y:

y=815.511 m This is the displacement  of the bullet after 6.9 s

8 0
3 years ago
A cyclist starts from rest and coasts down a 4.0∘ hill. The mass of the cyclist plus bicycle is 85 kg. Ignore air resistance and
Angelina_Jolie [31]

A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m

net work done by gravity on the cyclist = mass * gravity * height diff.

= 85 * 9.8 * 12.56

= 10470J

= 10.5kJ

B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J

vel.^2 = 10470 * 2 / 85 = 246.4

vel. = 15.7m/s

5 0
3 years ago
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