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3 years ago

8

Identify the element that has nine protons

1 answer:

loris [4]3 years ago

5 0

Fluorine has nine protons

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g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevell

olga55 [171]

Answer:

The value is $V = 2 V$

Explanation:

From the question we are told that

The length of the wire is $l = 10 \ m$

The current density is $J = 4*10^6 \ A/m^2$

The conductivity is $\sigma = 2*10^{7} \ S/m$

Generally conductivity is mathematically represented as

$\sigma = \frac{l}{RA}$

Here R is the resistance which is mathematically represented as

$R = \frac{V}{I}$

Here I is the current which is mathematically represented as

$I = J * A$

So

$R = \frac{V}{ J * A}$

And

$\sigma = \frac{l}{\frac{V}{ J * A} * A}$

=> $\sigma = \frac{l}{\frac{V}{J}}$

=> $V = \frac{l * J}{\sigma }$

=> $V = \frac{10 * 4*10^6}{2*10^{7} }$

=> $V = 2 V$

5 0

3 years ago

Newtons third lawWhat action-reaction forces are involved when a rocket engine fires? Why doesnt a rocket need air to push on? A

dangina [55]

Answer: action forc roketorce

reaction force is engine fires

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3 years ago

Physical values in the real world have two components: magnitude and

Nadusha1986 [10]

Answer:

Dimension.

Explanation:

6 0

3 years ago

In a magnetized substance, the domains __________________________ .

ahrayia [7]

I think the answer you're looking for is -are randomly oriented. if not sorry... i tried.

5 0

3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago