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3 years ago

Twin sisters Annie and Hannah are fighting

Physics

2 answers:

s344n2d4d5 [400]3 years ago

8 0

Fnet = -25N + 30N = 5N
(25N is negative since left is negative)

Papessa [141]3 years ago

3 0

Answer:

Fnet

Explanation:

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Una rueda gira con una frecuencia de 530 rpm. Determina la velocidad angular, el periodo y la frecuencia.

Vlad1618 [11]

Answer:

donde esta la bibliotekaaa

Explanation:

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6 0

2 years ago

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

The ball rotates 6.78 revolutions.

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

3 years ago

Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?

Assoli18 [71]

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion $s=ut+\frac{1}{2}at^2$

$40=0\times 10+\frac{1}{2}a10^2$

$a=0.8944m/sec^2$

Now from first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity, a is acceleration and t is time

So $v=u+at=0+0.8944\times 10=8.944m/sec$

6 0

3 years ago

The critical angle for a liquid in air is 520, What is the liquid's index of refraction? 0.62 0.79 1.27 1.50

sammy [17]

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, $\theta_c=52^o$

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

$n_1sin\theta_c=n_2sin\theta_2$

Here, $\theta_2=90$

$sin\theta_c=\dfrac{n_2}{n_1}$

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

$n_1=\dfrac{n_2}{sin\theta_c}$

$n_1=\dfrac{1}{sin(52)}$

n₁ = 1.269

or n₁ = 1.27

Hence, the refractive index of liquid is 1.27

8 0

3 years ago

Hi, I am having some difficulties in solving this question. Could someone please explain this question to me in detail. The thin

aniked [119]

Answer:

a) 3.0×10⁸ m

b) 0 m

Explanation:

Displacement is the distance from the starting position to the final position.

a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°). The distance between the points is the diameter of the circle.

x = 2r

x = 2 (1.5×10⁸ m)

x = 3.0×10⁸ m

b) In a full year, the Earth travels one full revolution, so it ends up back where it started. The displacement is therefore 0 m.

8 0

3 years ago