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3 years ago

How do physicists use science?

Physics

1 answer:

Ksju [112]3 years ago

6 0

Answer:B, They preform experiments.

Explanation:

your welcome :)

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The flowers of the bunchberry plant open with astonishing force and speed, with pollen grains accelerating from rest to 7.5 m/s.

zimovet [89]

Answer:

a) J= $7.5*10^{-10}$

b) F = $2.5*10^{-6}N$

c) The force is 2551.02 times bigger than the weight of the pollen grain.

Explanation:

a) we know that the impulse J is equal to:

J = $MV_f-MV_i$

where M is the mass, $V_f$ is the final velocity and $V_i$ the initial velocity. So, replacing values, we get:

J = $(10^{-10})(7.5m/s)-(10^{-10}(0)$

J = $(10^{-10})(7.5m/s)$

J= $7.5*10^{-10}$

b) We know that the force is equal to:

F = J/t

where J is the impulse and t the time. So, replacing values, we get:

F = $\frac{7.5*10^{-10}}{0.0003}$

F = $2.5*10^{-6}N$

c) The weight W of the pollen grain is calculated as:

W = $10^{-10}*9.8$

W = $9.8*10^{-10}N$

So, if we compare the force and the weight of the pollen grain, we get:

$\frac{2.5*10^{-6} }{9.8*10^{-10} } =2551.02$

it means that the force is 2551.02 times bigger than the weight of the pollen grain.

3 0

3 years ago

Read 2 more answers

Never change a subscript to balance an equation.<br> O<br> A. True<br> B. False

vlada-n [284]

Answer:

false

Explanation:

Subscript cannot be change to balance chemical equations.

7 0

3 years ago

Tired of those advertisers on brainly directing you towards buying premium?

Nadya [2.5K]

Answer:

Explanation:

5 0

4 years ago

How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?

RSB [31]

A small vehicle with less mass and with less Kinetic Energy will require less distance to stop than a large vehicle.

8 0

3 years ago

A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms

Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2$

The formula that is used to find the rms value of the electric field is as follows :

$I=\epsilon_o cE^2_{rms}$

c is speed of light and $\epsilon_o$ is permittivity of free space

So,

$E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m$

Hence, this is the required solution.

4 0

3 years ago