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frosja888 [35]
3 years ago
5

The half-life for a 100-gram sample of radioactive element X is 5 days. How much of element X remains after 10 days have passed?

Physics
1 answer:
Serga [27]3 years ago
3 0
Ok so this is a basic quest d
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What is one way we can use agriscience to affect photosynthesis and increase plant growth?
pochemuha

Answer:

casa. Completa las oraciones con el pronombre de objeto directo correcto, según el contexto.

1. Tu ropa está sucia. ¿Por qué no ____ lavas?

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5. ¿Quieres ir al cine con nosotros? _____ llamamos antes de salir.

6 0
3 years ago
what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec
s344n2d4d5 [400]
Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)


Formula (Kinetic energy)
E_{k} =  \frac{m*s^2}{2}

Solving:
E_{k} = \frac{m*s^2}{2}
E_{k} =  \frac{1*3^2}{2}
E_{k} =  \frac{1*9}{2}
E_{k} =  \frac{9}{2}
\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark
3 0
4 years ago
A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What
Harrizon [31]

The concepts used to solve this problem are those related to the Pythagorean theorem for which we will calculate the distance and the pitch.

According to the attached diagram we have that the expression of the resulting displacement is

R = \sqrt{(3b)^2+(4b)^2}

Therefore the resultant displacement of the girl is

R = \sqrt{(9b^2+16b^2)}

R = \sqrt{25b^2}

R = 5b

Therefore the girl has displaced around of 5 blocks

4 0
3 years ago
A toy car has a momentum of 3 kilogram meters per second south. The car has a 1-kilogram mass. Which is the velocity of the car?
antiseptic1488 [7]
Using p = mv 3 = 1× v v = 3m/s
6 0
3 years ago
Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up
Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

v = \sqrt{2as}

v = \sqrt{2*18.9*391}

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

V=  \sqrt{v^2 + 2aS'}

V = \sqrt{121.57^2+ 2 (-9.92)(332)}

V = 90.51 m/s

6 0
3 years ago
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