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3 years ago

The Northern Hemisphere is warmer in spring than in winter, because in spring

Physics

2 answers:

Allisa [31]3 years ago

6 0

Answer:

Both hemispheres are warmer in their Spring than they are

in their Winter, because . . .

A). the sun climbs higher in the sky during Spring than it does during

WInter ... shooting its rays more directly at the ground ...,

and

B). the sun stays up in the sky longer in Spring than it does in

Winter, giving the ground more time to absorb its rays.

Read more on Brainly.com - brainly.com/question/788447#readmore

Explanation:

Amanda [17]3 years ago

4 0

Both hemispheres are warmer in their Spring than they are
in their Winter, because . . .

A). the sun climbs higher in the sky during Spring than it does during
WInter ... shooting its rays more directly at the ground ...,

and

B). the sun stays up in the sky longer in Spring than it does in
Winter, giving the ground more time to absorb its rays.

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Gina made a poster for plastic recycling week and included this information on her poster:

miskamm [114]

Answer:

Gina should put “rubber tires” under “Synthetic.”

Gina should put “starch” under “Natural.”

Explanation: I just did the assignment ;P

4 0

3 years ago

Read 2 more answers

A MEMS-based accelerometer has a mass of m = 2 grams, an equivalent spring constant of k = 5 N/m, and an equivalent damping coef

pychu [463]

Answer:

The natural frequency = 50 rad/s = 7.96 Hz

Damping ratio = 0.5

Explanation:

The natural frequency is calculated in this manner

w = √(k/m)

k = spring constant = 5 N/m

m = mass = 2 g = 0.002 kg

w = √(5/0.002) = 50 rad/s

w = 2πf

50 = 2πf

f = 50/(2π) = 7.96 Hz

Damping ratio = c/[2√(mk)] = 0.1/(2 × √(5 × 0.002)) = 0.5

5 0

3 years ago

A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)

kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = $\frac{4\pi}{3}\times(0.0991^2-0.0882^2)$

volume of sphere = 0.0012 m³

also, volume of half sphere = $\frac{\textup{Total volume}}{\textup{2}}$

volume of half sphere = $\frac{\textup{0.0012}}{\textup{2}}$

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density = $\frac{\textup{Mass}}{\textup{Volume}}$

Density = $\frac{\textup{0.568}}{\textup{0.0012}}$

Density = 474 kg/m³

8 0

2 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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5 0

3 years ago

Two vectors, X and Y, form a right angle. Vector X is 48 inches long and vector Y is 14 inches long. The length of the

Andrews [41]

Answer:

The length of the resultant vector is 50 inches

Explanation:

Use the Pythagorean theorem to find the answer, since the addition of these two perpendicular vectors will have a magnitude (length) equal to the hypotenuse of the right angle triangle formed by the two:

$|Resultant|\,=\,\sqrt{x^2+y^2} \\|Resultant|\,=\,\sqrt{48^2+14^2}\\|Resultant|\,=\,\sqrt{2304+196}\\|Resultant|\,=\,\sqrt{2500}\\|Resultant|\,=\,50\,in$

The length of the resultant vector is 50 inches

3 0

3 years ago