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3 years ago

91.0x26x504 ...............

Physics

2 answers:

Svet_ta [14]3 years ago

8 0

please mark this answer as brainlist

Hoochie [10]3 years ago

7 0

Answer:

1192464

Explanation:

91.0 times 26 = 2366

2366 times 504 = 1192464

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What is the ultimate source of energy that makes the Sun shine?

zavuch27 [327]

Answer:

The sun's energy comes from thermonuclear fusion reactions.

Explanation:

Due to the Sun's strong gravitational pull, hydrogen atoms fuse, resulting in helium atoms. During this process, tremendous amounts of energy are released, or the energy of the Sun.

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3 years ago

The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one

ozzi

Answer:

e. 400 Hz

Explanation:

In closed organ pipe, only odd harmonics of fundamental note is possible .

The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .

200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc

600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .

Frequency not possible is 400 Hz .

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4 years ago

On the moon what would be the force of gravity acting on an object that has a mass of 7 kg?

Rudiy27

Answer:

Explanation:

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3 years ago

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

4 years ago

T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.

Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

$T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}$

Let me know if you have any questions.

7 0

3 years ago