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Sedbober [7]
3 years ago
10

Nuclei of u-238 atoms are

Physics
1 answer:
Elis [28]3 years ago
5 0
 <span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
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Ball A is negatively charged and ball B is electrically neutral. What happens to the charges of both balls when they come into c
tresset_1 [31]

Explanation:

They will repel, thus made of electrical conductor.

5 0
2 years ago
Normal faults occur from _____ stress and reverse faults occur from _____ stress.
natta225 [31]

Normal faults can be caused by physical stress. 
Reverse faults can be caused by emotional stress.

3 0
3 years ago
What is the upper temperature range for stars?
padilas [110]
The upper temperature range for stars is : c. 40,000 K

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This indicate the maximum possible energy that a star could release to its surrounding if its somehow exploded.
6 0
2 years ago
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A racing car has a mass of 750 kg. It undergoes an acceleration of 4.00 m/s2. What is the net force acting on the car?
navik [9.2K]

Answer:

<h2>3000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 4 = 3000

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<h3>3000 N</h3>

Hope this helps you

3 0
2 years ago
a) A space vehicle is launched vertically upward from the Earth's surface with an initial speed of 8.90 km/s, which is less than
elena-14-01-66 [18.8K]

Answer:

A) 4037.2[km]; B) 21472[m/s]

Explanation:

A)

This part can be solved using the principle of energy conservation, where kinetic energy will be equal to potential energy. We will define the kinetic energy and the potential energy.

Ek= 0.5*m*v^2\\where:\\m = mass [kg]\\v = 8.90*10^3[m/s]\\Ep=m*g*h\\where:\\g = gravity = 9.81[m/s^2]\\h = elevation or height [m]\\

Now we have to match both equations in this way the mass value is canceled and we can clear h.

0.5*m*(8.9*10^3)^{2}=m*9.81*h\\ h=4037206.93[m] = 4037.2[km]

B)

To solve this problem we can use the kinematic equations, but first we must identify the initial data:

yo = 2.35*10^7[m]

y = 0 [m] when hit the ground [m]

vo = 0 [m/s], It is essentially at rest...

v_{f} ^{2} =v_{o} ^{2} +2*a*(y)\\v_{f} =\sqrt{2*9.81*2.35*10^{7} } \\v_{f} =21472.54[m/s] o 21.47[km/s]

4 0
2 years ago
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