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schepotkina [342]
3 years ago
15

25. How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?

Physics
1 answer:
Lina20 [59]3 years ago
4 0
  • Length=l=4m

\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{l}{g}}

\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{4}{9.8}}

\\ \rm\rightarrowtail T=2\pi(0.63887)

\\ \rm\rightarrowtail T=1.27774\pi

\\ \rm\rightarrowtail T=4.012s

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A novelty golf ball of mass m is launched with an initial velocity v0 = (25i + 13j) m/s and then follows a parabolic trajectory.
SVEN [57.7K]

Explanation:

(a)

The initial vertical velocity is 13 m/s.  At the maximum height, the vertical velocity is 0 m/s.

v = at + v₀

0 = (-9.8) t + 13

t ≈ 1.33 s

(b)

Immediately prior to the explosion, the ball is at the maximum height.  Here, the vertical velocity is 0 m/s, and the horizontal velocity is constant at 25 m/s.

v = √(vx² + vy²)

v = √(25² + 0²)

v = 25 m/s

(c)

Momentum is conserved before and after the explosion.

In the x direction:

m vx = ma vax + mb vbx

m (25) = (⅓ m) (0) + (⅔ m) (vbx)

25m = (⅔ m) (vbx)

25 = ⅔ vbx

vbx = 37.5 m/s

And in the y direction:

m vy = ma vay + mb vby

m (0) = (⅓ m) (0) + (⅔ m) (vby)

0 = (⅔ m) (vby)

vby = 0 m/s

Since the vertical velocity hasn't changed, and since Fragment B lands at the same height it was launched from, it will have a vertical velocity equal in magnitude and opposite in direction as its initial velocity.

vy = -13 m/s

And the horizontal velocity will stay constant.

vx = 37.5 m/s

The velocity vector is (37.5 i - 13 j) m/s.  The magnitude is:

v = √(vx² + vy²)

v = √(37.5² + (-13)²)

v ≈ 39.7 m/s

7 0
4 years ago
Please help me! I'll give brainliest!!!!
poizon [28]

Answer:

Because they have already made an impact within our atmosphere

6 0
3 years ago
Read 2 more answers
A water-skier is moving at a speed of 14.3 m/s. When she skis in the same direction as a traveling wave, she springs upward ever
kiruha [24]

Answer:

a) 1.95 m/s

b) 5.56 m

Explanation:

Given that:

Velocity of the skier (V_s) = 14.3 m/s

For the skier moving in the direction of the wave, we have:

Period (T) = 0.450 s

Relative velocity (V) of the skier in regard with the wave =  (V_s - V_w)

where:

V_s = velocity of the skier

V_w = velocity of the wave

The wavelength (\lambda) can be written as:

\lambda = (V_s-V_w)T

\lambda = (V_s-V_w) 0.450m ---------------> Equation (1)

For the skier moving opposite in the direction of the wave, we have:

Period (T) = 0.342 s

Relative velocity (V) of the skier in regard with the wave = (V_s + V_w)

The wavelength (\lambda) can be written as:

\lambda = (V_s+V_w)T

\lambda = (V_s+V_w) 0.342m   ------------------> Equation 2

Equating equation (1) and equation (2) and substituting  V_s  = 14.3 m/s ; we have:

(V_s-V_w) 0.450m  =  (V_s-V_w) 0.342m

0.450m(V_s)-0.450m(V_w)   =  0.342m(V_s)+0.342m(V_w)

Collecting the like terms; we have:

0.450m(V_s) - 0.342m(V_s) =  0.342m(V_w)+0.450m(V_w)

(V_s)(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.108m =  (V_w)0.792m

1.5444m^2/s =  (V_w)0.792m

(V_w) = \frac{1.5444m^2/s}{ 0.792m}

(V_w) = 1.95 m/s

b)

The Wavelength of the wave can be calculated using :  ( \lambda }) = (V_s-V_w) 0.450m

({\lambda}) = (14.3 m/s -1.95 m/s)(0.450)

(\lambda) = (12.35)0.450m

(\lambda)= 5.5575 m

λ ≅ 5.56 m

5 0
3 years ago
In this video, the linear motion of the descending sphere is directly related to the rotational motion of the wheel, and in orde
matrenka [14]

Explanation:

Let a is the linear acceleration of the descending sphere. It is given by,

a=\dfrac{dv}{dt}.......(1) (change in velocity)

Let \alpha is the  angular acceleration α of the rotating wheel. It is given by :

\alpha =\dfrac{d\omega}{dt}............(2) (change in angular velocity)

Dividing equation (1) and (2) we get :

\dfrac{a}{\alpha }=\dfrac{dv}{d\omega}

Since, v=r\omega

\dfrac{a}{\alpha }=\dfrac{d(r\omega)}{d\omega}

\dfrac{a}{\alpha }=r

a=\alpha r

a=\alpha \times r

Hence, this is the required solution.

8 0
3 years ago
If the weight of an object on the moon is 1/6 of the weight of an object on earth. What
zubka84 [21]

Answer:

1200N

Explanation:

Suppose a body of mass "m" and its weight on the moon is Wm (where W is the weight and "m" is the moon ;which means weight on the moon).Mass of the moon is "M"

and its radius is "R"

Weight of an object on the moon = "F"(Force)with which the moon pulls.

Wm = GM*m/r2

Weight of the same object on the earth is We(where W is the weight and "e" is the earth; which means weight on the earth).

Mass of the earth is 100 times of that of the moon.

Radius of the moon = R

Radius of the Earth = 4R

Weight of the object on the moon =

We = G100M*m/(4R)2(Pronounced 4 R square)

We = G100M*m/(16R)2(Pronounced 16 R square)

Wm/We = G * M * m * 16R2/R2 * g * 100M * m

=16/100

Therefore, 4800N on earth= 1200N on moon

8 0
3 years ago
Read 2 more answers
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