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wolverine [178]
3 years ago
13

Why do you experience a shock or even see a spark when you reach for a doorknob.

Physics
2 answers:
Verizon [17]3 years ago
4 0

Answer: static electricity

Explanation:

You see, you rubbed your body either against a coat or the carpet. It send negative electrons throughout your body and once you hit metal with your hand it creates an electric shock that’s called static electricity.

tresset_1 [31]3 years ago
3 0
When you touch a doorknob, all the charge wants to leave you and go to the doorknob. You see a spark and get a shock as the electrons leave you. Lightning is the result of static electricity.
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A. <span> using a ramp to load boxes onto a truck </span>
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What's the relationship between magnetism and birds?​
neonofarm [45]

Answer:

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Explanation:

3 0
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What makes psychoanalysis different from behaviorism?
svp [43]
Psychoanalysis is a form of therapy that treats mental disorders through different ways, such as dream interpretation. It is a way of looking into the line between conscious and <span>unconsciousness </span>

Behaviorism is a theory that says that mental disorders are treated better with altering the persons behavior patterns.

I think they're different because behaviorism is just simply a theory and not actually a form of therapy.

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4 0
3 years ago
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
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