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sergiy2304 [10]
4 years ago
6

in an experiment which the volume of dry sand is measured by the displacement of water, the sand was slightly wet to begin with.

What effect would this have on the volume of air space that was calculated? on the percentage of the volume that was air space? This is from 8th grade introductory physical science book 6th edition.
Physics
1 answer:
Mamont248 [21]4 years ago
3 0
It would mean that you could not know the precise volume of the sand. Only the volume of the sand plus the water that was making it damp.

In the experiments listed, the effects are easy to deduce by understanding that the water in the sand adds volume to the 'sample' being measured.

So in the case of calculating air space you would calculate <em>less</em> air space.
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A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’
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A charged particle moving in a magnetic field experiences a force equal to:

\vec{F}=q\vec{v}\times \vec{B}

Thus, the magnitude of the force that the proton experiences is given by:

F=qvBsin\theta

The magnetic field is perpendicular to the proton's velocity, therefore, we have \theta=90^\circ. Replacing the given values, we obtain:

F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N

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3 years ago
2) As a freshman in college, you enter a classroom hall believing it is Physics 101 and sit down. The professor says, "During th
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No because you don’t learn about synthetic inventions yet in your first year
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3 years ago
You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it
VladimirAG [237]

Answer:

(A) The speed just as it left the ground is 30.25 m/s

(B) The maximum height of the rock is 46.69 m

Explanation:

Given;

weight of rock, w = mg  = 20 N

speed of the rock at 14.8 m, u = 25 m/s

(a) Apply work energy theorem to find its speed just as it left the ground

work = Δ kinetic energy

F x d = ¹/₂mv² - ¹/₂mu²

mg x d = ¹/₂m(v² - u²)

g x d = ¹/₂(v² - u²)

gd = ¹/₂(v² - u²)

2gd = v² - u²

v² = 2gd  + u²

v² = 2(9.8)(14.8) + (25)²

v² = 915.05

v = √915.05

v = 30.25 m/s

B) Use the work-energy theorem to find its maximum height

the initial velocity of the rock = 30.25 m/s

at maximum height, the final velocity = 0

- mg x H = ¹/₂mv² - ¹/₂mu²

- mg x H = ¹/₂m(0) - ¹/₂mu²

- mg x H = - ¹/₂mu²

2g x H = u²

H = u² / 2g

H = (30.25)² / 2(9.8)

H = 46.69 m

4 0
3 years ago
Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le
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Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = \frac{d}{2}

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

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