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3 years ago

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:

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

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

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

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

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

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

$(V_w) = 1.95 m/s$

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