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musickatia [10]
3 years ago
14

While running around the track at school, Milt notices that he runs due East in the 100m homestretch and due West on the 100m ba

ckstretch. To train for the 800 m event, he runs both the 100 m homestretch and the 100m backstretch in 12.2 seconds. How do Mills velocities on the homestretch and backstretch compare
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
5 0

They have equal magnitudes and opposite directions.

olga_2 [115]3 years ago
4 0

Answer:

The magnitude of both velocities will be same but direction will be opposite.

Explanation:

Distance on Homestretch = d1 = 100 m

Distance on backstretch = d2 = 100 m

Time = t = 12.2 s

Magnitude of velocity on Homestretch = V1 = d1/t = 100/12.2 = 8.196 m/s

Magnitude of velocity on backstretch = V2 = d2/t = 100/12.2 = 8.196 m/s  

if we look at the racecourse, we will come to know that runner runs on homestretch and backstretch in opposite direction. Hence, magnitude of velocities will be same but direction will be opposite.  

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What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

  • the quantity expressed by the body resisting angular acceleration.
  • It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

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7 0
2 years ago
Which instrument would make rice vibrate easier, a tuba or a flute? Explain why. Hint: think about the difference between high a
algol13

Answer:

I assume the higher notes would make the rice vibrate more easily, so a flute.

8 0
3 years ago
A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide
-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0
3 years ago
Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
deff fn [24]

Answer:

 electric field E = - k Q (1 /r(r-a)), force    F = - k Q qo / r (r-a) and force for r>>a    F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

         E = - k Q (1 /r(r-a))

b) We calculate the force.  

         F = E qo

         F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

         F ≈ - k Q qo / r²

6 0
3 years ago
Three bulbs are connected in series to a 4.5 V battery. The second bulb burns out. The
Oksi-84 [34.3K]

Answer:

0v

Explanation:

3 0
3 years ago
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