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Pepsi [2]
2 years ago
4

Sonya is playing a board game, and each space on the board game measures 1 centimeter. She moves her game token 5 spaces up from

the start position. Then she moves it 5 spaces to the left. Finally, Sonya moves her token 2 spaces down. What is the total distance the token moved?
Physics
2 answers:
lozanna [386]2 years ago
7 0

the total distance in all would be 12 spaces but if you where talking about in distance which you where it would be 3 centimeters from the starting line


DochEvi [55]2 years ago
4 0

Answer:

The total distance the token moved 12 spaces.

Explanation:

Given that,

Sonya is playing a board game. She moves her game token 5 space up from the start position and then she moves it 5 space to the left then she moves her token 2 space down.

Each space on the board game measure 1 centimeter.

We need to calculate the total distance the token moved

D= 5 space up from the start position + 5 space to the left + 2 space down

D=5+5+2

D=12\ spaces

Hence, The total distance the token moved 12 spaces.

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8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
A. Why is the stratosphere considered a "stable" layer in the atmosphere?
vovangra [49]

Answer:

A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.

B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.

C. The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .

D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.

E. Wind shear is the change in wind direction or speed with height in an atmosphere.

Explanation:

7 0
2 years ago
The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (I
FrozenT [24]

Answer:

The acceleration you can get with that engine in your car is around 70,56 (\frac{m}{s^{2} }) or 7,26 (\frac{ft}{s^{2} } ) using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

F=m*a

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }

a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }

<em> Note: N or Newton units are: \frac{kg * m}{s^{2} }</em>

a= 70,54 \frac{m}{s^{2} }

Also in pounds you can compared

a= \frac{2400  lf }{3 306  lf}

Note: lf in force units are: \frac{lf*ft}{s^{2} }

a=7,26 \frac{ft}{s^{2} }

8 0
2 years ago
Consider two adjacent states, S1 and S2, that wish to control particulate emissions from power plants and cement plants; New Jer
natka813 [3]

Answer:

a. 7500

b. Yes

c. 2500

d. 7500

Explanation:

Please see attachment

7 0
2 years ago
A radioactive source has a half life of 80s.
Burka [1]
Lets make the original number of nuclides at the start is 100.

If 7/8 of 100 is decayed, that means 87.5 decayed.

\frac{7}{8} \times 100 = 87.5

And there is 1/8 left of the number of nuclide 100. Which is 12.5

100 - 87.5 =12.5

\frac{1}{8} \times 100 = 12.5

How many Half lifes passed for 100 to become 12.5 is 3 Half-Lives.

100 \div 2 \div 2 \div 2 = 12.5

Each Half-Life is 80 seconds so there is 240 seconds

3 \times 80 = 240The answer is that it takes 240 seconds.
6 0
2 years ago
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