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3 years ago

6

Which of the following can influence one’s ability to detect stimuli and increase one’s responsiveness to stimuli in dangerous s

ituations?

2 answers:

Keith_Richards [23]3 years ago

6 0

The correct answer is Response Criteria.

nata0808 [166]3 years ago

6 0

Answer:

Response Criteria.

Explanation:

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Which Chinese man was leader of the National People's Party, called the Guomindang, in 1925?

Len [333]

<span>C. Mao Zedong
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5 0

3 years ago

Read 2 more answers

Mr. Phillips' car is parked on a steep hill with the brakes applied and the engine off. Because of the car's position, it has gr

aivan3 [116]

Answer:

The anser is b

Explanation:

4 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

I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT

ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR = E- V

IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

$R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms$

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0

3 years ago

At what speed does a 1400 compact car have the same kinetic energy as a 20000 truck going 25 ?

kumpel [21]

K=1400*V^2/2
K=20000*25^2/2. => 1400*V^2/2=20000*25^2/2 <=> 1400*V^2=20000*25^2
14*V^2=200*225
v^2=100*225/7
v=250/7^(1/2)

Answer: 250*7^(1/2)/7

4 0

3 years ago