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Law Incorporation [45]
4 years ago
14

The temperature in the upper atmosphere is -2°C, and the temperature on Earth’s surface is 5°C. Which type of weather is most li

kely in these conditions?
A.
freezing rain
B.
snow
C.
sleet
D.
rain
E.
drought
Physics
2 answers:
Aleks04 [339]4 years ago
8 0

A is the answer I think. :)

Ostrovityanka [42]4 years ago
8 0

Answer:

A. Freezing rain

Explanation:

Freezing rain is formed at a cold temperature below 0 degrees Celsius. Freezing rain falls at temperatures below freezing ( -2^{o}C ) by the ambient air mass that causes freezing on contact with Earth surface at about 5^{o} C.

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a runner moves 2.88 m/s north. she accelerates at 0.350 m/s2 at -52.0 angle. at the point where she is running directly east, wh
Naily [24]

Answer:

11.7 m

Explanation:

I assume north is the y direction and x is the east direction, so Δx refers to the displacement in the east direction.

First, find the time it takes for the velocity to change from directly north to directly east.

Given (in the y direction):

v₀ = 2.88 m/s

v = 0 m/s

a = 0.350 m/s² sin(-52.0°) = -0.276 m/s²

Find: t

v = at + v₀

(0 m/s) = (-0.276 m/s²) t + (2.88 m/s)

t = 10.4 s

Given (in the x direction):

v₀ = 0 m/s

a = 0.350 m/s² cos(-52.0°) = 0.215 m/s²

t = 10.4 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10.4 s) + ½ (0.215 m/s²) (10.4 s)²

Δx = 11.7 m

7 0
3 years ago
There is a dropping of a coconut of 2kg mass. the speed of the coconut at the moment before it hits the ground is 40m s-1.
igor_vitrenko [27]
  • Mass=2kg
  • Velocity=40m/s

We know

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

  • m denotes to mass
  • v denotes to velocity

\\ \sf\longmapsto K.E=\dfrac{1}{2}\times 2\times 40^2

\\ \sf\longmapsto K.E=40^2

\\ \sf\longmapsto K.E=1600J

6 0
3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
True or false?<br>nuclear fusion happens when uranium combines ​
Arturiano [62]

Answer: false

Explanation:

.

5 0
3 years ago
Read 2 more answers
The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given  

N = W Cos30

N = (500) Cos30

N = 433 N

3 0
3 years ago
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