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2 years ago

pumili Ng dalawang teoryang napag aralan gumawa Ng Venn diagram na nag papakita Ng pag kakatulad at pagkaiba Ng mga ito

Physics

1 answer:

MrRissso [65]2 years ago

7 0

Answer:

you will be my girl my my girl myyyyy girllll you will be my girl my girl myyyy worldddddd you will be my. smokin ciggarets on teh roof you look so pretty and i love this view we fell inlove in october and thats why i love fall

Explanation:

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A baseball diamond is a square (don’t be fooled that it is usually shown rotated by 45°). Each side of the square is 90 feet lon

gayaneshka [121]

A line from second to home creates a right triangle with 45 degree angles.

This makes the diagonal line the side length of the square multiplied by the sqrt(2)

Second to home = 90sqrt(2) = 127.279 feet ( round answer as needed)

8 0

2 years ago

After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

klio [65]

1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

Explanation:

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

$p_1=(2)(0)=0$

Using the same equation used in part 1), the initial momentum of the bowling ball is

$p=mv$

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

$p_2=(8)(20)=160 kg m/s$

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

$p_i = p_1 + p_2$

where

$p_1$ is the initial momentum of the trophy

$p_2$ is the initial momentum of the ball

Here we have

$p_1 = 0$

$p_2 = 160 kg m/s$

Therefore, the total momentum is

$p_i = 0 + 160 = 160 kg m/s$

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

$p_i = p_f$

where

$p_i$ is the total momentum before the collision

$p_f$ is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

$p_f = 160 kg m/s$

We can write the total momentum after the collision as

$p_f = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the trophy

$v_1$ is the final velocity of the trophy

$m_2 = 8 kg$ is the mass of the bowling ball

$v_2 = 12 m/s$ is the final velocity of the ball

Since we also know the value of the final total momentum,

$p_f = 160 kg m/s$

we can solve the equation to find the velocity of the trophy:

$v_1 = \frac{p_f - m_2 v_2 }{m_1}=\frac{160-(8)(12)}{2}=32 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

4 0

3 years ago

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m

lesya [120]

Answer:

a) A =0.021525m

b) $\phi=0.37869rad$

c) $v_{max}=5.4098\frac{m}{s}$

d) $y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

Explanation:

1) Notation

A= Amplitude

v= velocity

$\lambda$ = wavelength

k= wave number

$\omega$ = angular frequency

f= frequency

2) Part a and b

The equation of movement for a transverse sinusoidal wave is gyben by (1)

$y(t)=Acos(kx+ \omega t +\phi)$ (1)

At x=0 ,t=0 we have that:

$0.02=Acos(\phi)$

The velocity would be the derivate of the position, so taking the derivate of (1) respect to t we got (2)

$v(t)=-\omega Asin(kx+ \omega t+\phi)$ (2)

And replacing the conditions at x=0, t=0 we got

$-2\frac{m}{s}=-\omega Asin(\phi)$

Now we can find the angular frequency with equation (3)

$\omega =\frac{2\pi}{T}$ (3)

Replacing the values obtained we got:

$\omega =\frac{2\pi}{0.025s}=80\pi \frac{rad}{s}$

From equation (1) we have:

$Acos(\phi)=0.02$ (a)

$-2=-80\pi Asin(\phi)$ (b)

So from condition (b) we have:

$Asin(\phi)=\frac{1}{40\pi}$ (c)

If we divide condition (c) by condition (a) we got:

$\frac{Asin(\phi)}{Acos(\phi)}=tan(\phi)=\frac{1}{0.02x40\pi}=\frac{1}{0.8\pi}=0.39789$

If we solve for $\phi$ we got:

$\phi =tan^{-1}(0.39789)=0.37869$

And now since we have $\phi$ we can find A from equation (a)

$Acos(0.37869)=0.02$

So then Solving for A we got $A=\frac{0.02}{cos(0.37869)}=0.021525$

3) Part c

From equation (2) we can see that the maximum speed occurs when $sin(\omega t+\phi)=1$ , so on this case we have:

$v_{max}=\omega A=80\pi \frac{rad}{s}x0.021525m=5.4098\frac{m}{s}$

4) Part d

On this case we need an equation like (1), and we have everything except the wave number, and we can obtain this from the following expression:

$v=\lambda f=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}$ (4)

And solving for k from equation (4) we got

$k=\frac{\omega}{v}=\frac{80\pi \frac{rad}{s}}{30\frac{m}{s}}=\frac{8\pi}{3}m^{-1}}$

And with the k number we have everythin in order to create the wave function, given by:

$y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

7 0

3 years ago

Pls someone help me pls…. and pls explain to me how

slavikrds [6]

Answer:

1.12 × 10⁴ m/s

Explanation:

The escape velocity of the object v = √(2GM/R) where G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of the Earth = 6 × 10²⁴ kg and R = radius of the Earth = 6.4 × 10⁶ m

Since v = √(2GM/R)

Substituting the values of the variables into the equation, we have

v = √(2GM/R)

v = √(2 × 6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)

v = √(13.34 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg/6.4 × 10⁶ m)

v = √(80.04 × 10⁻¹¹ × 10²⁴Nm²/kg/6.4 × 10⁶ m)

v = √(80.04 × 10¹³Nm²/kg ÷ 6.4 × 10⁶ m)

v = √(80.04 ÷ 6.4 × 10¹³ ÷ 10⁶Nm/kg)

v = √(12.50625 × 10⁷ Nm/kg)

v = √(125.0625 × 10⁶ Nm/kg)

v = 11.18 × 10³ m/s

v = 1.118 × 10 × 10³ m/s

v = 1.118 × 10⁴ m/s

v ≅ 1.12 × 10⁴ m/s

8 0

2 years ago

A horizontal spring with spring constant 85 N/mN/m extends outward from a wall just above floor level. A 4.5 kgkg box sliding ac

swat32

Answer:

v = 0.028 m/s

Explanation:

In this case you take into account that all the elastic potential of the box, when the spring is compressed is equal to the kinetic energy of the box before it hits the spring. That is:

$K=U$