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guajiro [1.7K]
3 years ago
8

An isotope has the same ________, but a different ________ .

Physics
2 answers:
11111nata11111 [884]3 years ago
6 0
I<span>sotopes of an element have the same number of protons but different number of neutrons.</span>
Lunna [17]3 years ago
4 0
An isotope is one of the forms of an element that has the same number of protons, but a different number of neutrons.
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Longitudional waves travel through a series of ________ and ___________.
nekit [7.7K]

Answer:

compressions; rarefactions

Explanation:

4 0
2 years ago
If a sound wave traveled from a solid to a liquid its speed would:.
JulijaS [17]

Answer:

increase

Please Give Brainliest!

5 0
2 years ago
Find the net force for <br> 10 N<br> 10 N<br> 25degree<br> 5N<br> 5N
icang [17]

Answer:

30n

Explanation:

7 0
2 years ago
The fulcrum of a uniform 20-kg seesaw that is 4.0 m long is located 2.5 m from one end. A 30-kg child sits on the long end. Part
uranmaximum [27]

Answer:

57 kg

Explanation:

Mass of seesaw = 20 kg

Length of seesaw = 4 m

Mass of child on the longer end = 30 kg

The weight of the seesaw acts at the center i.e. 2m

The algebraic sum of moments of all forces about any point is zero, hence, using the fulcrum as the reference point:

[x * 9.8* 1.5] - [20 * 9.8* (2.5 - 2)] - [30 * 9.8 * 2.5] = 0

=> 14.7x = (20*9.8*0.5) + 735

14.7x = 98 + 735

14.7x = 833

=> x = 833/14.7

x = 57 kg

7 0
3 years ago
In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape
lina2011 [118]

Answer:

The deceleration of the pilot was  1.9 x 10⁴ m/s²

Explanation:

First, let´s calculate the velocity of the pilot 3 m above the ground. To do that, we need to know how much time the pilot was falling. The equation for the position and velocity of the pilot are as follows:

y = y0 + v0 * t + 1/2 * g * t²

v = v0 + g * t

where

y = position of the pilot at time t

y0 = initial position

v0 = initial velocity

t = time

v = velocity at time t

g = acceleration due to gravity

If we consider the ground as the center of our reference system and that the pilot fell with an initial velocity of 0 (the pilot would unlikely impulse himself to the ground), then the equations would be as follows:

y = y0 + 1/2 g * t²  

v = g * t

The time at which the pilot was 3.0 m above the ground will be:

3.0 m = 6000 m - 1/2 * 9.8 m/s² * t²

3.0 m - 6000 m / -4.9 m/s² = t²

t = 35.0 s

The velocity at that time will be:

v = -9.8 m/s² * 35.0 s = -343 m/s

After 35.0 s the pilot has a positive acceleration besides the acceleration due to gravity. Then, the equation for velocity and position will be:

v = v0 + a*t       now, v0 = -343 m/s and a ≠ g and a>0

y = y0 + v0 * t + 1/2 * a * t²    now, y0 = 3 m

Again, let´s find the time at which the pilot hits the ground:

v = v0 + a*t  

v-v0/ t  = a

Replacing in the equation for position:

y = y0 + v0 * t + 1/2 * ((v-v0)/t) * t²

y = y0 + v0 * t + 1/2 * v* t - 1/2 * v0 * t)

y = y0 + 1/2 v0 * t + 1/2 v * t

replacing with numbers:

0m = 3m + 1/2 * (-343 m/s)t + 1/2 *(-54 m/s)t

-3.0 m = - 198.5 m/s * t

t = -3m / -198.5 m/s

t =0.015 s

the upward acceleration was then:

v-v0/ t  = a

-54 m/s -(-343 m/s) / 0.015 s = a

a = 1.9 x 10⁴ m/s²

6 0
3 years ago
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