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Licemer1 [7]
2 years ago
9

A 5.0 kg book is lying on a 0.25 meter high table.

Physics
1 answer:
Wittaler [7]2 years ago
3 0

Answer:

Below in the picture:-

I hope it helps.

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A cobalt atom contains 27 protons, 32 neutrons, and 27 electrons. What is
kirza4 [7]
32? I could be wrong but I’m going with that answer choice
3 0
2 years ago
If the resistance of a circuit increases, that means that the current has to
Gnoma [55]

Answer:

If resistance increases current decreases.

Explanation:

  • Current is <em>inversely proportional</em> to the resistance.
  • from the relation given below, we can clearly see the relation between current and resistance;

                              V=IR

                              I ∝ 1/R

This relation shows that when resistance increases,current decreases.

4 0
3 years ago
Systemic circulation requires help from which of the following bones, lungs, arteries and veins, or muscles
ahrayia [7]
Systemic circulation is between the heart and the rest of the body. Therefore arteries and veins are needed.
4 0
3 years ago
A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
3 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

t = 387.69 yrs

5 0
3 years ago
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