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ch4aika [34]
3 years ago
8

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

35 m. If he completes the 200 m dash in 30.0 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track? m/s2
Physics
1 answer:
Pie3 years ago
5 0

Answer

given,

distance = 200 m

radius of curvature = 35 m

time = 30 s

centripetal acceleration = ?

speed of the runner

v = \dfrac{d}{t}

v = \dfrac{200}{35}

v = 5.71 m/s

acceleration of the runner

a_c = \dfrac{v^2}{r}

a_c = \dfrac{5.71^2}{35}

a_c = 0.931 m/s^2

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alukav5142 [94]
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Quantum numbers allow us to dig deeper into the electron configurations by allowing us to focus on electrons' quantum nature. This includes such properties as principle energy (size) (n), magnitude of angular momentum (shape) (l), orientation in space (m), and the spinning nature of the electron. In terms of connecting quantum numbers back to electron configurations, n is related to the energy level, l is related to the subshell, m is related to the orbital, and s is due to Pauli Exclusion Principle.</span>
7 0
3 years ago
A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0
3 years ago
Weight of a man is less at coal mine.Why?
dmitriy555 [2]
If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.
5 0
3 years ago
Read 2 more answers
A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?
Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒  Density = \frac{Mass}{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

On substituting the values, we get

⇒               =\frac{6.3 \ g}{19.3 \ g/mL}

⇒               =0.32 \ mL

7 0
3 years ago
A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 3.84 m/s
JulijaS [17]

Answer:

the train is moving at the speed of v = 1.79 m/s

Explanation:

given,

rain drop is falling vertically down with the speed of  = 3.84 m/s

angle of the rain drop = 25°                      

tan θ = \dfrac{v}{u}                      

tan 25° = \dfrac{v}{3.84}                      

v =3.84 × tan 25°                      

v = 1.79 m/s                  

hence, the train is moving at the speed of v = 1.79 m/s

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3 years ago
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