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allsm [11]
3 years ago
7

It takes brooke 10minutes to run 1 mile what is her speed in miles per minutes

Physics
1 answer:
hjlf3 years ago
6 0

If you divide miles by minutes, the answer will have units of
miles per minute, which is exactly what you want.

(1 mile) / (10 minutes) = 1/10  mile/minute = 0.1 mile per minute


You might be interested in
A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where
4vir4ik [10]

Answer:

Explanation:

Inductance L = 1.4 x 10⁻³ H

Capacitance C = 1 x 10⁻⁶ F

a )

current I = 14 .0 t

dI / dt  = 14

voltage across inductor

= L dI / dt

= 1.4 x 10⁻³ x 14

= 19.6 x 10⁻³ V

= 19.6 mV

It does not depend upon time because it is constant at 19.6 mV.

b )

Voltage across capacitor

V = ∫ dq / C

= 1 / C ∫ I dt  

= 1 / C ∫ 14 t dt

1 / C x 14 t² / 2

= 7 t² / C

= 7 t² / 1 x 10⁻⁶

c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance

energy stored in inductor

= 1/2 L I²

energy stored in capacitor

= 1/2 CV²

After time t

1/2 L I² = 1/2 CV²

L I² =  CV²

L x ( 14 t )² = C x  ( 7 t² / C )²

L x 196 t² = 49 t⁴ / C

t² = CL x 196 / 49

t = 74.8 μ s

After 74.8 μ s energy stored in capacitor exceeds that of inductor.

7 0
3 years ago
Unpolarized light with an intensity of 22.4 ????ux passes through a polarizer whose transmission axis is vertically oriented. (a
irina1246 [14]

Answer:

a)      I₁ = 11.2 Lux , vertical direction , b)      I₂ = 1.44 Lux

Explanation:

a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed

consequently the above the processed light has half of the incident intensity and the directional of the polarized

          I₁ = I₀ / 2

          I₁ = 22.4 / 2

          I₁ = 11.2 Lux

is polarized in the vertical direction

b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus

         I₂ = I₁ cos² θ

         I₂ = 11.2 cos² 69

         I₂ = 1.44 Lux

8 0
3 years ago
The wavelengths for visible light rays correspond to which of these options?
ryzh [129]

Answer:

sorry about the other person but its b

Explanation:

5 0
2 years ago
A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
What type of acceleration does an object moving with constant speed in a circular path experience?Select one:A) constant acceler
olga_2 [115]

ANSWER:

D) centripetal acceleration.

STEP-BY-STEP EXPLANATION:

When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.

Therefore, the answer is centripetal acceleration.

3 0
1 year ago
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