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allochka39001 [22]
3 years ago
12

Radon has a half-life of 56s. After what time will it's activity be reduced to (a) 90% (b) 50% (c) 10% of it's initial value​

Physics
1 answer:
guapka [62]3 years ago
6 0

Answer:

A) 11.2 s

B) is 56 s

C) 170.8 s

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Calcular la densidad de un trozo de hierro cuya masa es 110 g<br>y ocupa un volumen de 13.99 cm?.​
polet [3.4K]

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I can't understand your question sorry

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How do you put 0.0015kg in scientific notation
Anna11 [10]
0.0015 kg would be 1.5*10^-2 kg in scientific notation


4 0
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NEED HELP NOW!!! I WILL BRAINLIEST.
Sergio039 [100]

Answer:

1. Carbon-12

2. Nitrogen-13

3. Carbon-13

7 0
3 years ago
A tennis ball moving at a speed of 4 m/s, collides with a bowling ball at rest. The tennis ball bounces back in the direction op
Alex Ar [27]

Answer:

Both experienced the same magnitude impulse

Explanation:

This is because, the impulse force is internal to the system of both the tennis ball and the bowling ball. It is an action-reaction pair. So, the force exerted on the tennis ball by the bowling ball equals in magnitude to the force exerted by the tennis ball on the bowling ball although, they are in opposite directions. This, both experienced the same magnitude impulse.

3 0
3 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

7 0
3 years ago
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