All categories

3 years ago

Since the Sun has more mass, why do objects on earth not move closer to the Sun instead of staying put on Earth?

Physics

1 answer:

maw [93]3 years ago

6 0

Answer:

Because the Earth has it's own gravity that keeps us put, and we also have the moon.

Explanation:

You might be interested in

The human nerve cells have a net negative charge and the material in the interior of the cell is a good conductor. if the cell h

mash [69]

The magnitude and direction (inward or outward) of the net flux through the cell boundary is - 0.887 wb.m².

Flux describes any effect that appears to pass or travel through a surface or substance.

The magnitude and direction (inward or outward) of the net flux through the cell boundary is calculated as follows;

Ф = Q/ε

where;

Q is net charge
ε is permittivity of free space

Φ = (-7.85 x 10⁻¹²)/(8.85 x 10⁻¹²)

Φ = - 0.887 wb.m²

Learn more about flux here: brainly.com/question/10736183

#SPJ1

6 0

2 years ago

How can resistance be useful and where can they be useful?

sergiy2304 [10]

-- Resistance can be useful among the population of a repressive government.
Although it can be dangerous for those who resist, it can also exert pressure
against the regime to alter its repressive practices.

-- Resistance can also be useful in electronic circuits. "Lumped" components with
known numerical values of resistance are used to divide voltage, limit current, and
dissipate controlled amounts of electrical energy.

3 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

What is transferred by a radio wave?

Alex

Answer: B. energy

Explanation:

4 0

3 years ago

An electric car is being designed to have an average power output of 4,600 W for 2 h before needing to be recharged. (Assume the

MAVERICK [17]

Answer:

a)3312 x 10⁴ J

b)I = 57.5 A

c)9200 W

Explanation:

Given that

P =4600 W

Time t= 2 h = 2 x 3600 s= 7200 s

We know that

1 W = 1 J/s

Energy stored in the battery = P .t

=4600 x 7200 J

=3312 x 10⁴ J

We know that power P given as

P = V .I

V=Voltage ,I =Current

4600 = 80 x I

I = 57.5 A

The energy supplied = 4600 x 2 = 9200 W

7 0

3 years ago