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vivado [14]
3 years ago
6

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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Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil
tatyana61 [14]

Answer:

The time taken by missile's clock is 4.6\times 10^{6} s

Solution:

As per the question:

Speed of the missile, v_{m = 6.5\times 10^{3}} m/s

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}         (1)

Using binomial theorem in the above eqn:

We know that:

(1 + x)^{a} = 1 + ax

Thus eqn (1) becomes:

1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}

T = \frac{2c^{2}}{v_{m}^{2}}

Now, putting appropriate values in the above eqn:

T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}

T = 4.6\times 10^{6} s

4 0
3 years ago
What is the potential energy for 65kg climber on top of Mount Everest (8,800 m high)
Fed [463]

Answer:

PE= m * g *h

work:

PE= 65kg * 9.8 kg *8,800 m

PE=5605600 m/kg

idk the actual units i forgot

6 0
2 years ago
A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha
lord [1]

Answer:

The value is F_f =  46.935 \  N

Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

       F_{net} =  F -  F_f =  ma

Here F_f is force of kinetic friction (in N) acting on the crate

      So  

            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

5 0
2 years ago
What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

5 0
3 years ago
What is the equivalent resistance for a series circuit with three resistors : 5.0 ohms, 2.0 ohms, and 12.0 ohms
vesna_86 [32]

Answer:19ohms

Explanation:

equivalent resistance=5+2+12

equivalent resistance=19ohms

8 0
3 years ago
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