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2 years ago

Which of the following statements correctly describes the movement of water?

Physics

2 answers:

Schach [20]2 years ago

8 0

More water evaporates than falls as rain
did u know 3 million tonnes of water gets evaporated from oceans per day

Llana [10]2 years ago

3 0

The answer would be D. More water evaporates than falls as rain.

You might be interested in

The dipole moment of HI is 0.42D. What is the dipole moment of HI in C⋅m?

sleet_krkn [62]

So in your question where ask to find the dipole moment of HI in C.M. So in my calculation by converting the given data from Debyes to Coulomb Meteres is that 1 Dyebes is equals to 3.33X10^(-30) C.M and the answer would be 1.40X10(-30)C.M. I hope you are satisfied with my answer and feel free to ask for more

5 0

3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

A charged particle of mass m = 5.00g and charge q = -70.0μC moves horizontally to the right at a constant velocity of v = 30.0 k

pychu [463]

Answer:

The magnitude of the force, B = 5 Tesla, Up (North) direction

Explanation:

Magnetic force F= Eq where Electric field, E = 750 NC

and charge, q = -70 μC = -7 × $10^{-5}$ C

F = 750 × -7 × $10^{-5}$

F = 0.0525

But F = qvB; B = $\frac{F}{qv}$

where B is the magnetic field

= 0.0525 ÷ ( -7 × $10^{-5}$ × 30)

B = 5.0 Teslas

The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Hence the direction of the charge is up (North).

8 0

2 years ago

Read 2 more answers

Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 200

SashulF [63]

Answer:

400m

Explanation:

Brainliest? :))

Let your initial displacement from your home to the store be

1 and your displacement from the store to your friend’s house

be Dd

Given: Dd

1 = 200 m [N]; Dd

2 = 600 m [S]

Required: Dd

Analysis: Dd

T 5 Dd

1 1 Dd

Solution: Figure 6 shows the given vectors, with the tip of Dd

joined to the tail of Dd

2. The resultant vector Dd

T is drawn in red,

from the tail of Dd

1 to the tip of Dd

2. The direction of Dd

T is [S].

T measures 4 cm in length in Figure 6, so using the scale of

1 cm : 100 m, the actual magnitude of Dd

T is 400 m.

Statement: Relative to your starting point at your home, your

total displacement is 400 m [S].

6 0

2 years ago

the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,

balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0

2 years ago