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navik [9.2K]
3 years ago
12

Radioactive materials have ununstable what

Physics
1 answer:
Zigmanuir [339]3 years ago
4 0
Radioactive material have unstable nuclei
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A 2kg baseball has a momentum of 75 kg m/s. Find it’s speed
almond37 [142]

Answer:

37.5m/s

Explanation:

momentum = mass x velocity

momentum / mass = velocity = speed

75kg*m/s / 2kg = 37.5m/s

6 0
3 years ago
The only drawback to solar energy is that it
MA_775_DIABLO [31]
It is to highly priced for most of civilization.
3 0
4 years ago
What is the total translational kinetic energy in a test chamber filled with nitrogen (N2) at 2.02×105 Pa and 20.7°C? The dimens
Mumz [18]

Answer:

K. E = 6.0796 × 10^ -21 Joule

Explanation: see attachment

3 0
4 years ago
on earth a block is placed on a frictionless table. When a 50 North horizontal force is applied to the block, it accelerates at
melamori03 [73]

As per Newton's II law we know that

F = ma

here

F = applied unbalanced force

m = mass of object

a = acceleration of object

now it is given that force F = 50 N North applied on block on earth due to which block will accelerate by 4 m/s^2

so here from above equation

50 = m* 4

m = \frac{50}{4} = 12.5 kg

Now we took another situation where block is placed on surface of moon and again force F = 25 N is applied on the block

So we will again use Newton's II law

F = ma

25 = 12.5 * a

a = \frac{25}{12.5}

a = 2 m/s^2

so block will accelerate on moon by acceleration 2 m/s^2

5 0
3 years ago
A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show
Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]

And the speed can be calculated as follows:

v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]

7 0
3 years ago
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