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3 years ago

Which event best helped becquerel determine uranium radiate rays?

1 answer:

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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A gun with a muzzle velocity of 100 m/s is fired horizontally from a tower. Neglecting air resistance, how far downrange will th

Neporo4naja [7]

Answer:

313.6 m downward

Explanation:

The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.

In fact, we have:

$y(t) = h +u_y t + \frac{1}{2}at^2$

where

y(t) is the vertical position of the projectile at time t

h is the initial height of the projectile

$u_y = 0$ is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally

t is the time

a = g = -9.8 m/s^2 is the acceleration due to gravity

We can rewrite the equation as

$y(t)-h = \frac{1}{2}gt^2$

where the term on the left, $y(t)-h$ , represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we find

$y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m$

So the bullet has travelled 313.6 m downward.

6 0

4 years ago

What is a phone message?

malfutka [58]

A message telephone messagemessage a communication usually brie that is written or spoken or signaled; "hesent a three-word <span>message"</span>

8 0

3 years ago

Read 2 more answers

Is the answer D?

irga5000 [103]

Nope answer is 500 N downward
draw the freebody diagram to understand

7 0

4 years ago

Describe how the resistance of the filament lamp changes as the current through it increases.

pochemuha

This heats up when an electric current passes through it, and produces light as a result. The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.

8 0

3 years ago

Assuming this is a distance time graph( ignore the speed time title) assume metres on vertical scale. describe in as much detail

riadik2000 [5.3K]

Answer:

The journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance travelled during the journey from the start point A to the final point B is 40 m

Explanation:

From the start point A to point B, we have;

The speed from A to B = 10 m/(10 s) = 1 m/s

The distance traveled from A to B = 10 m

The time it takes to move from A to B = 10 seconds

From the point B to point C, we have;

The distance traveled from B to C = 0 m, (stationary)

The time it remains at point B distance from the start point = 10 seconds

The speed between point B to C = 0 m/(10 s) = 0 m/s

From the point C to point D, we have;

The distance traveled from C to D = 10 m

The time it takes to move from C to D = 5 seconds

The speed between point C and D = 10 m/(5 s) = 2 m/s

From the point D to point E, we have;

The distance traveled from D to E = 0 m, (stationary)

The time it remains at point D distance from the start point = 10 seconds

The speed between point D to E = 0 m/(10 s) = 0 m/s

From the point E to point F, we have;

The distance traveled from E to F = 20 m (return journey starts at point E)

The time it takes to move from E to F = 5 seconds

The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)

Therefore, the journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;

d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m

The total distance travelled in the journey is 40 m

The total displacement, $\underset{d}{\rightarrow}$ = 10 m + 10 m - 20 m = 0 m

7 0

3 years ago