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alex41 [277]
2 years ago
7

What method could I use to test this hypothesis? If the mass and the volume of and object are known, then its density can be cal

culated dividing the object's mass by its volume.
Physics
1 answer:
Nitella [24]2 years ago
7 0

Answer:

The scientific method

Explanation:

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Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of
Lana71 [14]

Electric field due to a point charge is given as

E = \frac{kq}{r^2}

here we know that

q = 1.6 \times 10^{-19} C

also the distance is given as

r = 5.29 \times 10^{-11} m

now we will have

E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}

so we will have

E = 5.14 \times 10^{11} N/C

so above is the electric field due to proton

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A 1.5 kg ball pushed with a force of 13.5 N accelerates to the left. What is the acceleration of the ball?​
GarryVolchara [31]

Explanation:

F= ma

13.5 =1.5a

1.5a/1.5 =13.5/1.5

a= 9 m/s^2

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if the train is accelerating and the bisicle is traveling at a constant velocity, what do you know about their speed?
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The train is accelerating meaning there is a change in the velocity so the speed is either increasing or decreasing depending. The bicycle is travelling at a constant velocity meaning it is travelling at a constant speed.
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Read 2 more answers
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
3 years ago
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