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Ainat [17]
3 years ago
7

A veterinarian thinks that a dog has swallowed a key ring. Which types of electromagnetic waves is the doctor most likely

Physics
2 answers:
poizon [28]3 years ago
4 0

Answer:

X rays

Explanation:

ryzh [129]3 years ago
3 0

Answer:

X-Ray

Explanation:

I took the test

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The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a)
Lunna [17]

Answer:

a

The number of radians turned by the wheel in 2s is   \theta= 8\ radians

b

The angular acceleration is  \alpha =14 rad/s^2

Explanation:

        The angular velocity  is given as

                 w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3

Now generally the integral of angular velocity gives angular displacement

           So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

    This is mathematically evaluated as

            \theta(t ) = \int\limits^2_0 {2t + t^3} \, dt

                  = [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.

                  = [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0

                  = 4 +4

                 \theta= 8\ radians

Now generally the derivative  of angular velocity gives angular acceleration

      So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

    This is mathematically evaluated as          

           \frac{dw}{dt}  = \alpha (t) = 2 + 3t^2

so at t=2

            \alpha (2) = 2 +3(2)^2

                   \alpha =14 rad/s^2

7 0
3 years ago
Which sentence describe newton's third law?
strojnjashka [21]
C. Forces are always in pairs
3 0
3 years ago
Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce
ASHA 777 [7]

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

F=k\frac{q_{1}q_{2}  }{r^{2} }

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

So

F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}  

4 0
4 years ago
In your own words explain why the thickness and length of a wire could effect the flow of electrons​
agasfer [191]

Answer:

(you can use my exact words) The length and thickness would make it so that the electrons move differently than they would a shorter and thinner wire because with the wire being longer the electrons would have a longer trip and with the wire being thicker the electrons would be more spread out and move be able to move more freely

8 0
3 years ago
Express the measurement 0.00000575 into scientific notation.
g100num [7]

Answer: = 5.75 × 10 -6

Explanation:

= 5.75 × 10-6

(scientific notation)

= 5.75e-6

(scientific e notation)

= 5.75 × 10-6

(engineering notation)

(millionth; prefix micro- (u))

= 0.00000575

(real number)

7 0
3 years ago
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