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blsea [12.9K]
2 years ago
13

Which event best helped becquerel determine uranium radiate rays?

Physics
1 answer:
agasfer [191]2 years ago
6 0
Exposure of a Photographic plate to Uranium in a closed space along with a metal that prevented the rays from passing through best helped becquerel determine Uranium radiates rays.
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2 years ago
You drop a ball at 27 feet high. How long will it take for the ball to hit the ground?
ivolga24 [154]

Ball will hit the ground after a time of 1.296 s

Explanation:

initial velocity of ball= Vi=0

g= 9.8 m/s²

height =h= 27 ft=8.23 m

using the kinematic equation

h= Vi t + 1/2 gt²

8.23=0(t) + 1/2 (9.8)t²

t²=1.6796

t=1.296 s

6 0
2 years ago
HELP PLEASE
denis-greek [22]
B) Distance: the farther, the weaker!
4 0
3 years ago
Read 2 more answers
Force (N) = 250<br> Mass (kg) = 221<br> Acceleration (m/s^2) = ?
lyudmila [28]

Answer:

The required acceleration is a=1.1 m/s²

Explanation:

Given

  • Force F = 250 N
  • Mass m = 221 kg

To determine

Acceleration a = ?

We know that acceleration is produced when a force is applied to a body.

The acceleration can be determined using the formula

F = ma

where

  • F is the force
  • m is the mass
  • a is the acceleration

now substituting F = 250 , and m = 221 in the formula

F = ma

250 = 221 (a)

switch the equation

221(a) = 250

Divide both sides by 221

\frac{221a}{221}=\frac{250}{221}

simplify

a=\frac{250}{221}

a=1.1 m/s²

Therefore, the required acceleration is a=1.1 m/s²

8 0
2 years ago
For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the maximum ra
umka2103 [35]
Refer to the diagram shown below.

The following discussion assumes a simplistic analysis that ignores air resistance and variations in the terrain that the missile travels over.

Let the launch velocity be V₀ at an angle of θ relative to the horizontal.
The horizontal component of velocity is V₀ cosθ.
If the time of flight is t_{f}, then
r=V_{o} \, t_{f}
where r = the range of the missile.

Also, the time, t, when the missile is at ground level is given by
0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2}
where g = acceleration due to gravity.

t = 0 corresponds to when the missile is launched. Therefore
t_{f} =  \frac{2V_{o}sin\theta}{g}

Therefore
r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g}

Typically, θ=45° to achieve maximum range, so that
r= \frac{V_{o}^{2}}{g}

This analysis is more applicable to a scud missile rather than a powered, guided missile.

Answer:
t_{f} =  \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g}
Usually, θ=45°

6 0
2 years ago
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