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2 years ago

5

Using up resources at a rate in which they can be replaced is called:

1 answer:

mash [69]2 years ago

7 0

Answer:

D it think

Explanation:

Not 100% sure .. Plzz don't feel offended if wrong

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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

2 years ago

Solids diffuse because the particles cannot move.

Nastasia [14]

Solids cannot diffuse.

4 0

2 years ago

Read 2 more answers

A larger object is made of two balls separated by a massless rigid rod that is 1.5 m long. The mass of the red ball at one end i

lions [1.4K]

Answer:

The speed of the 1 kg red ball 8.04 m/s .

Explanation:

Given :

Separation between rods , d = 1.5 m .

Mass of the red ball is 1 kg .

Mass of the orange ball is 5.7 kg .

Angular velocity , $\omega=1\ rev/s = 2\pi\ rad/s$ .

Now , distance of center of mass from red ball is :

$r =\dfrac{0\times 1+1.5\times 5.7}{1+5.7}\\\\r = 1.28\ m$

We know , speed is given by :

$v=\omega r\\\\v=2 \times \pi \times 1.28\\\\v=8.04\ m/s$

Hence , this is the required solution .

5 0

2 years ago

Select the correct answer. A ball is thrown straight down from the top of a building at a velocity of 16 ft/s. The building is 4

BartSMP [9]

Answer:

The ball takes 5s to reach the ground

Explanation:

in order to solve this problem we use the kinematics equation with gravity as acceleration:

$h=v_0t+1/2*gt^2$

we replace the values

$0=1/2*32t^2+16t-480$

We solve this quadratic equation:

t=5s

t=-6s (this solution has not physical sense)

8 0

2 years ago

a radioactive element has a half-life of 30 days. calculate the mass of the element that remains after 20 days if 100g of the el

patriot [66]

Explanation:

the answer is in the image above

4 0

2 years ago