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3 years ago

11

The range of temperatures on earth is much less than the range in temperatures on mercury, in part due to the presence of oceans

on earthtrue or false

1 answer:

Feliz [49]3 years ago

8 0

False, the real reason is the inconsistency of how far away the sun is

You might be interested in

Which items in this image are electrically conductive? Check all that apply.

balu736 [363]

Answer: a and d

Explanation: A.) the power lines themselves

B.) the wooden pole that supports the lines

C.) the rubber soles on the worker’s boots

D.) the metal tools the worker uses

E.) the wooden ladder leaning against the lines

4 0

3 years ago

A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a

GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

$T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}$

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

$T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}$

T = 7.83 X10⁻⁷ s

6 0

3 years ago

Which Box in the diagram above, should this picture go into?

lyudmila [28]

What diagram? There isn’t one

7 0

3 years ago

The current through a 10-ohm resistor connected to a 120-V power supply is The current through a 10-ohm resistor connected to a

Mazyrski [523]

Answer:

12 A

Explanation:

V= I * R

V/R = I

120 V /10 Ω = 12 AMPS

7 0

2 years ago

A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is

Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d= $\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m$

Wavelength of light, $\lambda=510 nm=510\times 10^{-9} m$

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

$sin\theta=\frac{m\lambda}{d}$

For first bright fringe, =1

$sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305$

$\theta=sin^{-1}(0.305)=17.76^{\circ}$

The distance between two m=1 fringes

$x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m$

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

$sin\theta=1$

$sin\theta=\frac{m\lambda}{d}$

Substitute the values

$1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}$

$m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3$

Maximum number of bright fringes on the scree= $2m+1=2(3)+1=7$

8 0

4 years ago