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3 years ago

How important friction is?

Physics

1 answer:

kirill [66]3 years ago

3 0

Friction is important because;

It enables us to walk due to the grip our feet have on the ground.
It enables the planes to fly due to the drag force.
It enables vehicles to move and to stop due to brakes and tire marks
Friction enables us to write using a pen.

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Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 kg Jan

Brut [27]

Complete part of Question: What is Jane's (and the vine's) angular speed just before she grabs Tarzan

Answer:

Jane's (and the vine's) angular speed just before she grabs Tarzan, w = 1.267 rad/s

Explanation:

According to the law of energy conservation:

Total change in kinetic energy = Total change in potential energy

Mass of Jane = 60 kg

Mass of the vine = 32 kg

Mass of Tarzan = 72 kg

Height of Tarzan = 5.50 m

Length of the vine = 8.50 m

Jane's change in gravitational potential energy,

$U_J = 60 * 9.8 * 5.5\\U_J = 3234 J$

Vine's gravitational potential energy,

$U_v = Mgh/2\\U_v = 32*9.8*5.5/2\\U_v = 862.4J$

Vine's Kinetic energy :

$KE_V = 0.5 I w^{2} \\I_V = \frac{ML^2}{3} = \frac{32 * 8.5^2}{3} = 770.67 kg m^2\\ KE_V = 0.5 *770.67 * w^{2}\\KE_V = 385.33 w^{2}$

Jane's Kinetic energy:

$KE_J = 0.5m(wL)^2\\KE_J = 0.5*60(w * 8.5)^2\\KE_J = 2167.5 w^2$

$U_J + U_V = KE_J + KE_V$

3234 + 862.4 = 2167.5w² + 385.33w²

4096.4 = 2552.83w²

w² = 4096.4/2552.83

w² = 1.605

w = √1.605

w = 1.267 rad/s

5 0

2 years ago

A body of mass 500kg moving at a speed of 10m/s reaches the speed of 50m/s in 20s.The force exerted is

Leona [35]

Answer:

answer is 1000 N

formula used-

F= m x (v-u/t)

mark this as brainliest!

6 0

2 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

2 years ago

A thin glass rod is submerged in oil. What is the critical angle for light traveling inside the rod? The index of refraction for

Alekssandra [29.7K]

When light travels from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle of incidence above which light is reflected only (no refraction occurs), and the value of this critical angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In this problem, the first medium is the glass ( $n_1 = 1.50$ ), while the second medium is oil ( $n_2 =1.46$ ), therefore the critical angle is given by
$\theta_c = \arcsin( \frac{1.46}{1.50} )=\arcsin(0.973)=76.7^{\circ}$

7 0

3 years ago

How many leptons are in ⁷₃Li?

shutvik [7]

-- Electrons are leptons. There are three electrons in each neutral Lithium atom.

The last two parts of the question are absurd.

-- Bonbons are candy, not atomic particles. A bonbon cannot fit into a Lithium atom.

-- A pentagon is a closed geometric figure that has five sides. Although you could, in principle, have a pentagon small enough to fit into a Lithium atom, you could never find a piece of paper small enough to draw it on.

5 0

2 years ago