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mart [117]
3 years ago
10

If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c

an you drive on a horizontal roadway around a right turn of radius 28.0 m before the cup starts to slide?
Physics
1 answer:
Cloud [144]3 years ago
3 0

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

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an alligator crawls 25m to the left with an average velocity of -1.2m/s. how many seconds did the alligator crawl?
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3 years ago
In the formula for magnetic deflection, FM is equal to qvB. Match each symbol in this formula with what it represents.
slava [35]

Answer:

1.F_m-Deflection force

2.q-Charge

3.v-Velocity of the charge

4.B-Perpendicular magnetic field strength

Explanation:

We are given that magnetic  deflection

F_m=qvB

We have to match each symbol in formula with what it  represents.

In this formula

F_m=Deflection force

q=Charge of particle

v=Velocity of charge

B=Perpendicular magnetic field strength

Therefore,

1.F_m-Deflection force

2.q-Charge

3.v-Velocity of the charge

4.B-Perpendicular magnetic field strength

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Scientists take scientific measurements carefully in order to ensure that reliability and validity. What is the difference betwe
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7 0
3 years ago
consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti
ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$\frac{Energy}{Area} = \sigma\times T^{4}

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$E= \sigma\times T^{4}

Now -

For Star A →

T_{A} = 4900 K

For Star B →

T_{B} = 9900 K

Therefore -

$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}

\frac{T_{B} }{T_{A} }= 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }

E(B) = E(A) x (\frac{T(B)}{T(A)} )^{4}

E(B) = E(A) x 2^{4}

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

brainly.com/question/13451162

#SPJ4

4 0
2 years ago
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