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2 years ago

Describe two ways force and motion are at work as you make the trip from your house to school

Physics

1 answer:

Juliette [100K]2 years ago

5 0

Answer:

If your driving, the car is moving from your home to school, so the car is in motion. Using the pedals on a car, which gives it gas or the breaks, is force because it keeps the car moving, and or can stop it.

Explanation:

Motion is when an object moves from one place to another, while force is what causes an object to move or to stop moving.

I hope this helps and that I answered you question correctly!

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A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

2 years ago

A glass window is coated with a transparent film of refractive index n2 =1.25. This film will cause destructive interference in

valina [46]

Answer:

0.00000011 m

Explanation:

n = 0 for minimum thickness

t = Thickness

$n_2$ = Refractive index of the film = 1.25

$\lambda$ = Wavelength = $550\times 10^{-9}\ m$

We have the formula

$2n_2t=(2n+1)\dfrac{\lambda}{2}$

$\\\Rightarrow t=(2n+1)\dfrac{\lambda}{4n_2}\\\Rightarrow t=(2\times 0+1)\dfrac{550\times 10^{-9}}{4\times 1.25}\\\Rightarrow t=0.00000011\ m$

The minimum thickness of the film is 0.00000011 m

6 0

2 years ago

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t

Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2 $\frac{m}{s^{2} }$

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

$K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=F$ Δx= $(0.400m)\\$

W_{ext}=20.4 Nm

$U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2} \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}$

To determine, the acceleration we solve the following equation for a:

$F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }$

8 0

2 years ago

A grinding wheel is spinning with an initial angular velocity +ω0. When its motor is turned off at t=0, it begins to slow down w

Musya8 [376]

Answer:

Explanation:

Given

initial angular velocity is $\omega _0$

when motor is turned of it started decelerating with $-\alpha$ angular acceleration

angle turned before coming to halt

$\omega ^2-\omega _0^2=2(\alpha )\theta$

here final angular velocity is zero

$-\omega _0^2=-2\alpha \theta$

$\theta =\frac{\omega _0^2}{2\alpha }$

No of revolutions will be $N=\frac{\theta }{2\pi }$

$N=\frac{\omega _0^2}{4\pi \alpha }$

5 0

2 years ago

What is Newton's second law of motion

Alexandra [31]

Newton’s second law is that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

8 0

2 years ago