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Savatey [412]
3 years ago
12

Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble

m. Car A uses tires for which the coefficient of static friction is 0.335 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 26.8 m/s. Car B uses tires for which the coefficient of static friction is 0.683 on the same curve. What is the maximum speed at which car B can negotiate the curve?
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v²  = μRg

= .683  x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

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A microwave oven has 1200 W. If it receives 120 V of potential difference, what is the current in the microwave?
8_murik_8 [283]

Answer:

10amps

Explanation:

Given data

Power =1200W

Voltage =120V

Required

The current I

From

P=IV

I= P/V

Substitute

I=1200/120

I= 10amps

Hence the current is 10amps

7 0
2 years ago
The length of the minute hand of the clock is 14cm. Calculate the speed with which the tip of the minute hand moves
Molodets [167]

Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

\omega= \frac{2\pi}{T}

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

8 0
3 years ago
What is the fate of solar radiation that reaches the earth?
expeople1 [14]
When solar radiation reaches the Earth it quickly dissipates as most of the radiation and UV rays are blocked by ozone layer, but more radiation and UV rays are able to get through because of global warming.
6 0
3 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
If Fg=mg solve for g
Bas_tet [7]

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

6 0
3 years ago
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