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sukhopar [10]
3 years ago
11

During radioactive decay, ___ energy was transformed to ____

Physics
1 answer:
Afina-wow [57]3 years ago
6 0

Answer:

Potential energy is transfer to kinetic energy,

If radio active substance is decayed.

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What happens to the electric potential energy of a negatively charged ion as it moves through the water from the negative probe
Nataly [62]

Answer:

Decreases.

Explanation:

Electric potential energy is the potential energy which is associated with the configuration of points charge in a system and it is the result of conservative coulomb force.

When the negatively charge ion is at the position of the negative probe than its potential energy is positive when it is move towards the positive probe it's potential energy becomes negative due to the negative ion.

Therefore, potential energy is decreases when negative charge ion moves through the water from negative probe to positive probe.

5 0
3 years ago
A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.
Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we  can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

y=(u sin \theta)t-\frac{1}{2}gt^2 (1)

where:

u sin \theta is the initial vertical velocity of the shell, with u=156 ft/s and \theta=49.0^{\circ}

g=32 ft/s^2 is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

x=(ucos \theta)t

where u cos \theta is the initial horizontal velocity of the shell.

We can re-write this last equation as

t=\frac{x}{u cos \theta} (1b)

And substituting into (1),

y=xtan\theta -\frac{1}{2}gt^2 (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of \alpha=-41.0^{\circ} from the horizontal, so we can write the position (x,y) of the hill as

y=x tan \alpha (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

xtan\alpha = xtan \theta - \frac{1}{2}gt^2

Substituting (1b) into this equation,

xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0

Which has 2 solutions:

x = 0 (origin)

and

tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft

So, the distance d down the hill at which the shell strikes the hill is

d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

t=\frac{x}{u cos \theta}

where:

x = 1322 ft is the final position of the shell when it strikes the hill

u=156 ft/s is the initial velocity of the shell

\theta=49.0^{\circ} is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s

Instead, the vertical component of the velocity is given by

v_y=usin \theta -gt

And substituting at t = 12.9 s (time at which the shell strikes the hill),

v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s

Therefore, the  final speed of the shell is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
Suppose a greenhouse is to be constructed to keep plants warm in the winter. Which material should be used for the windows of th
Anastasy [175]

Answer:

the answers, material D meets the requested characteristics

Explanation:

The objective of an insulating material for the house, must allow solar radiation to enter, so that the plants can perform photosynthesis and must prevent radiation from inside the house from being lost.

Therefore the material must meet two conditions be transparent to sunlight and be absorbed from the radiation coming from the house; this is to leave for visible light and absorb infrared radiation

Reviewing the answers, material D meets the requested characteristics

4 0
2 years ago
If you drag a sled on the ground, what does the sled experience?
Ilya [14]
The sled experiences friction against the ground
3 0
3 years ago
A student is trying to decide what to wear. His bedroom is at 20.0°C. His skin temperature is 30.0°C. The area of his exposed sk
Advocard [28]

Answer:

Explanation:

Stefan's Boltzman Law gives the value of radiation from a black body at a particular temperature. The relation is

E = eσ ( T₂⁴ -T₁⁴ )

e is emmisivity , σ is a constant = 5.67 x 10⁻⁸ , E is energy emitted per unit time per unit area

E = .915 x 5.67 x 10⁻⁸ x ( 303⁴ -293⁴ )

= .915 x 5.67 x 10⁻⁸ x ( 84.28 -73.70 ) x 10⁸

= 55 W / m²s

Area A = 1.45 m²

time t = 14.3 min

= 858 s

Total radiation

= 55 x 858 x 1.45 W

= 68425 J

5 0
3 years ago
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