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allsm [11]
3 years ago
10

Bipolar disorder, major depression, and anxiety disorders may all be treated with

Physics
2 answers:
olga nikolaevna [1]3 years ago
7 0
Anti anxiety medications
tangare [24]3 years ago
3 0

Answer:

Bipolar disorder, major depression, and anxiety disorders may all be treated with  anti-depressant medications.

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Are electrons lower in mass than neutrons
tamaranim1 [39]

Answer:

ya I just looked it up on Google

3 0
3 years ago
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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.55 times a second. A tack is stuck in the tire a
earnstyle [38]

Answer:

Tangential speed = 5.72 m/s

Centripetal acceleration = 91.6\text{ m/s}{}^2

Explanation:

The tangential speed, V, is given by

v=\omega r

where \omega is the angular speed and is given by 2\pi f (f is the angular frequency or frequency of rotation)

Thus,

v=2\pi f r = 2\times3.14\times2.55\times0.357 = 5.72\text{ m/s}

The centripetal acceleration,a, is given by

a=\dfrac{v^2}{r}

a=\dfrac{5.72^2}{0.357} = 91.6\text{ m/s}{}^2

7 0
4 years ago
Read 2 more answers
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
In which atmospheric layer are almost all water-based clouds formed?
luda_lava [24]

Answer and Explanation:

In troposphere layer all water based cloud formed. Troposphere is the lowest layer of the atmosphere. The troposphere is slightly unstable layer the weather occurs mainly in troposphere the most cloud formed in this layer troposphere is closed to the earth its depth varied according to the regions of the earth. Troposphere is more dipper in warmer region .

8 0
3 years ago
Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Eart
Temka [501]

The distance from the Earth's center to the point outside the Earth is 55800 Km

<h3>How to determine the distance from the surface of the Earth</h3>
  • Acceleration due to gravity of Earth = 9.8 m/s²
  • Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²
  • Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
  • Mass of the Earth (M) = 5.97×10²⁴ Kg
  • Distance from the surface of the Earth (r) =?

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by g

r² = GM / g

Take the square root of both sides

r = √(GM / g)

r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]

r = 4.94×10⁷ m

Divide by 1000 to express in Km

r = 4.94×10⁷ / 1000

r = 4.94×10⁴ Km

<h3>How to determine the distance from the center of the Earth</h3>
  • Distance from the surface of the Earth (r) = 4.94×10⁴ Km
  • Radius of the Earth (R) = 6400 Km
  • Distance from the centre of the Earth =?

Distance from the centre of the Earth = R + r

Distance from the centre of the Earth = 6400 + 4.94×10⁴

Distance from the centre of the Earth = 55800 Km

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

5 0
2 years ago
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