Ask question

Ask question

All categories

3 years ago

10

Which of the following statements is considered weather? It is very hot in the desert. It snows a lot in Antarctica. The sun is

always shining in Arizona. There is a tornado in Oklahoma right now.

1 answer:

azamat3 years ago

4 0

It is very hot in the desert

You might be interested in

Complete the following sentences.

MrMuchimi

force pull the ball towards it with a force called gravity. <em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>
newton

5 0

3 years ago

A very long stick ruled with meter markings is placed in empty space. A spaceship of rest length L = 100 m runs lengthwise along

alexira [117]

Answer:

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m

Explanation:

Given data:

Spaceship length of L = 100 m

Relative velocity between the ship and stick is given as

$v = \frac{4}{5} c$

The observed length observed by the outside observer is

$L' = L\sqrt{1 -\frac{v^2}{c^2}}$

putting all value to get observe length

put $v = \frac{4}{5} c$

$L ' = 100\times \sqrt{1 - \frac{(4/5)^2c^2}{c^2}$

L' = 60 m

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m

8 0

3 years ago

Helppppppppppppppppppppppppp

andrew11 [14]

Answer:

value of x=35°

y=72.5°

hope it helps you

make me brainliest plz

6 0

3 years ago

A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j

fenix001 [56]

Answer:

$F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k$

Explanation:

When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both

so first the force on the moving charge due to electric field is given by

$\vec F_e = q\vec E$

$\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3$

$\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j$

Now force on moving charge due to magnetic field is given as

$\vec F_b = q(\vec v \times \vec B)$

$\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )$

$\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i$

$\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k$

Now net force due to both

$F = F_e + F_b$

$F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k$

7 0

3 years ago

Calculate the total energy of 4.0 kg object moving horizontally at 20 m/s 50 meters above the surface.

Serhud [2]

Answer:

Correct answer: E total = 2,800 J

Explanation:

Given:

m = 4 kg the mass of the object

V = 20 m/s the speed (velocity) of the object

H = 50 m the height of the object above the surface

E total = ? J

The total energy of an object is equal to the sum of potential and kinetic energy

E total = Ep + Ek

Ep = m g H we take g = 10 m/s²

Ep = 4 · 10 · 50 = 2,000 J

Ek = m V² / 2

Ek = 4 · 20² / 2 = 2 · 400 = 800 J

E total = 2,000 + 800 = 2,800 J

E total = 2,800 J

God is with you!!!

4 0

3 years ago