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Maru [420]
3 years ago
10

What are 5 hazards of static electricity?​

Physics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

It is dangerous when there are flammable gases or a high concentration of oxygen. A spark could ignite the gases and cause an explosion. It is dangerous when you touch something with a large electric charge on it. The charge will flow through your body causing an electric shock.

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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
2 years ago
What color could be created by mixing cyan and yellow pigment?
Cloud [144]

Answer:

green pigment

Explanation:

green pigment

3 0
2 years ago
Read 2 more answers
NEED HELP ASAP WILL MARK BRAINLIEST.
-Dominant- [34]
G is the answer for apex vs / Chehhhh
3 0
2 years ago
A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the
lilavasa [31]

Answer:

R=803k\Omega

Explanation:

We have the following information,

V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s

We apply the equation for capacitor charging the voltage across it,

V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}

Replacing values,

R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}

R=803k\Omega

3 0
3 years ago
Brainliest please help<br><br>tell me if am right <br>if not correct me <br><br><br>​please
REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

6=2+a*2\\4=2*a\\a=2[m/s^{2} ]

<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

25=5+2*t\\t = 10 [s]

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

v_{f}=4+10*2\\v_{f}=24 [m/s]

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

v_{f}=5+8*10\\v_{f}=85 [m/s]

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0

8 0
3 years ago
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