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3 years ago

Which describes why radioisotopes can be used as tracers in studying the biological and chemical processes of plants?

Physics

1 answer:

netineya [11]3 years ago

8 0

Answer:

The radioisotope is chemically identical to the nonradioisotope the plant normally uses.

Explanation:

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A train moves at a constant velocity of 90 km/h. How far will it move in 0.25h?<br>

Serjik [45]

In 0.25h it will move in 22.5 kilometers.

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2 years ago

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What is the momentum of a 2.0kg ball rolling at 6.0 m/s

irga5000 [103]

Momentum can be defined by the formula p=m*V (where m is mass and V is velocity) so if we plug in these numbers:

p = 2kg * 6m/s
p = 12 kgm/s

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3 years ago

Acceleration that causes objects to move in a circle is called .

murzikaleks [220]

Centripetal Acceleration

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2 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

8 0

2 years ago

g Determine the magnitude of the electric field at the surface of a lead-208 nucleus, which contains 82 protons and 126 neutrons

Nookie1986 [14]

Answer:

Explanation:

Electric field at the surface of the the lead 208 = KQ/ R²

where K = 8.99 × 10⁹ Nm² /C²

Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C

V of the lead = 208 v of a proton assuming they both are sphere

4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton

R³ = 208 r³

R = ∛( 208 r³) = 5.92r

replace r with 1.20 x 10-15 m

R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m

E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² = 0.233 × 10²² N/C = 2.33 × 10²¹ N/C

4 0

2 years ago