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saw5 [17]
2 years ago
6

A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards

is the positive velocity direction.
A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,0) and (1,5). This first line is connected to a second line with endpoints (1,5) and (3,-5). This second line is then connected to a third line of endpoints (3,-5) and (6,-5).




A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)". A curve that relates v to t is shown in blue. It begins with a straight line of endpoints (0,0) and (1,5). This first line is connected to a second line with endpoints (1,5) and (3,-5). This second line is then connected to a third line of endpoints (3,-5) and (6,-5).
What is the badger's displacement \Delta xΔxdelta, x from t=2\,\text st=2st, equals, 2, start text, s, end text to 3\,\text s3s3, start text, s, end text?
Physics
1 answer:
Mrrafil [7]2 years ago
5 0

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

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charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel
IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁  + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³  N / C .

3 0
2 years ago
Is v2 = v1t+a dimensionally correct? Explain please!
Lady bird [3.3K]
You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities. 
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
 the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s). 
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.

4 0
2 years ago
What is the frequency of radiation whose wavelength is 11.5 a0 ?
irakobra [83]

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

4 0
2 years ago
Many birds can attain very high speeds when diving. Using radar, scientists measured the altitude of a barn swallow in a vertica
scoray [572]

Answer:

0.109

Explanation:

8 0
2 years ago
A TV satellite broadcasts at a frequency of 5000 MHz, (1 MHz = 1 million Hertz). What is the wavelength of this radiation?
Nitella [24]

Answer:

\lambda=0.06\ m

Explanation:

Given:

  • frequency of the broadcast, f=5000\ MHz=5\times 10^9\ Hz
  • we have the speed of the radiation equal to the speed of light, c=3\times 10^8\ m.s^{-1}

The broadcast waves are the electromagnetic waves but it can travel only upto a hundred kilometers without any loss of information carried by it.

<u>The relation between the frequency and the wavelength:</u>

\lambda=\frac{c}{f}

\lambda=\frac{3\times 10^8}{5\times 10^9}

\lambda=0.06\ m

5 0
2 years ago
Read 2 more answers
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