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Mazyrski [523]
3 years ago
5

The most important difference between a wedge and an inclined plane is ____.

Physics
1 answer:
podryga [215]3 years ago
8 0

C-how they are used. An incline plane is used as a ramp and a wedge is used in between something.

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4. A disobedient student dropped his Physics textbook (mass 0.1kg) from the window (15m above the ground). How fast was it going
vaieri [72.5K]

Answer:

v= 17.15 m/s

Explanation:

mass of the book=0.1 Kg

height above ground, h= 15 m

Using conservation of energy

Potential energy is converted into kinetic energy

mgh = \frac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 15}

v=\sqrt{294}

v= 17.15 m/s

Hence, the book will hit the ground at the speed of 17.15 m/s.

8 0
2 years ago
Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite
Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass_{Hank × Acceleration_{Hank = Mass_{Henry × Acceleration_{Henry

so

Mass_{Hank /  Mass_{Henry = Acceleration_{Henry / Acceleration_{Hank

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass_{Hank /  Mass_{Henry = 1 / 1.26

Mass_{Hank /  Mass_{Henry = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0
3 years ago
What happens in a tug of war if the net forces are balanced and why?
FinnZ [79.3K]

Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

6 0
4 years ago
A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

$\mathrm{u}=\cos (\omega \mathrm{t})

Then:

$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0
2 years ago
The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small c
Elodia [21]

Answer:

the distance between adjacent fringes is increased by a factor o 2

Explanation:

To find how the distance between fringes is modified you can use the following formula for the calculation of the distance between fringes:

\Delta y=\frac{\lambda D}{d}

D: distance to the screen

d: distance between slits

λ: wavelength of the light

if d is decreased by a factor of 2, that is d'=1/2d, you have:

\Delta y'=\frac{\lambda D}{d'}=\frac{\lambda D}{(1/2)d}=2\Delta y

hence, the distance between adjacent fringes is increased by a factor o 2

4 0
3 years ago
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