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Dahasolnce [82]
2 years ago
12

Karen is running forward at a speed of 9 m/s. She tosses her sweaty headband backward at a speed of 20 m/s. The speed of the hea

dband, to the nearest whole number relative to a stationary observer watching Karen, is _____ m/s
Physics
2 answers:
Komok [63]2 years ago
8 0
Let Karen's forward speed be considered as positive.
Therefore, before the headband is tossed backward, the speed of the headband is
V = 9 m/s

The headband is tossed backward relative to Karen at a speed of 20 m/s. Therefore the speed of the headband relative to Karen is
U = -20 m/s

The absolute speed of the headband, relative to a stationary observer is
V - U
= 9 + (-20)
= - 11 m/s

Answer:
The stationary observes the headband traveling (in the opposite direction to Karen) at a speed of 11 m/s backward.

Advocard [28]2 years ago
7 0

Answer:

-11 m/s

Explanation:

a lot simpler than that above me

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A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this
noname [10]

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

W=Fdcos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

The displacement of the object is d = 3.00 m, while \theta=180^{\circ}, because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J

And the work done is negative, because it is done against the motion of the object.


6 0
2 years ago
Read 2 more answers
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude
bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

\omega_{f}=0.240+0.917\times0.203

\omega_{f}=0.426\ rev/s

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of  tangential speed

v= r\omega

Put the value into the formula

v = \dfrac{0.720 }{2}\times0.426\times2\pi

v=0.963\ m/s

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.36\times0.917\times2\pi

a_{c}=2.074\ m/s^2

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0
2 years ago
An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and
serg [7]
<h2>The flux through the infinite charged wire along the central axis of a cylindrical surface of radius r and length l is  ∅E = E x 2πrl   </h2>

Explanation:

let us consider a thin infinitely long straight wire having a uniform charge density λ Cm⁻¹.To determine the field at a distance r from the line charge , we have cylindrical gaussian surface of radius r, length l,and with its axis along the line charge. it has curved surface S₁ , and flat circular ends S₂ and S₃. Obviously, dS₁//E, dS₂ ⊥E , and dS₃ ⊥ E , so, only the curved surface contributes towards the total flux.

  ∅E = ∫ E.dS = ∫E.dS₁ +∫E.dS₂ +∫E.dS₃

        = ∫EdS₁ cos0⁰ +∫EdS₂ cos 90⁰ +∫Eds₃ cos 90⁰

        = E∫ds₁₁ +0+0

         = E x area of curved surface

     ∅E = E x 2πrl    

3 0
2 years ago
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0
2 years ago
At which latitudes shown in the image of Earth do people experience the greatest tangential speed? Explain why.
sashaice [31]

Tangential speed is dependent on an object's distance from the center of the circle. The farther away from the center an object is, the faster the object has to travel. Therefore, people living at a latitude of 0 degrees experience the greatest tangential speed.



7 0
2 years ago
Read 2 more answers
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