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egoroff_w [7]
3 years ago
11

Two masses (5.3kg and 7.5kg) are fastened together with a small amount of explosive. They are loaded into a spring gun that is t

ilted 60.0 ◦ above the horizontal and launched at a speed of 25m/s. At the top of the trajectory, the explosive detonates (assume the mass of explosive is negligble). After it detonates, the 5.3kg mass isn’t moving at all. What is the velocity of the second mass just after the explosion?
Physics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

8.2

Explanation:

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If a closed series circuit has a voltage of 12 V and R1 = 4 Ω and R2 = 2 Ω, what is the current running through R1 and R2?
Phantasy [73]
Total resistance=R1+ R2= 6Ω
Voltage=12v
Current =\frac{voltage}{resistance}
Current= 2A
In a series circuit, equal current passes through every resistance.
Answer is option A
7 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}}

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5 0
3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3
bagirrra123 [75]

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

v = 117.22 m/s

Approximation to 2 sig. figures gives;

v = 120 m/s

8 0
3 years ago
4. Interference is an example of which aspect of electromagnetic radiation?
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D is the answer wave behavior

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