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3 years ago

Friction is always undesirable. True False

Physics

2 answers:

Anna007 [38]3 years ago

3 0

False because we use friction for many things . thus it is useful

LuckyWell [14K]3 years ago

3 0

this answer would be false

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Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her

leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$

7 0

2 years ago

You run a race with your friend. At first you each have the same kinetic energy, but then you find that she is beating you. When

Blizzard [7]

Answer:

52.49 Kg

Explanation:

Let m1 and v1 denote your mass and velocity respectively

Let m2 and v2 denote your friends mass and velocity respectively

Kinetic energy is given by

$KE= 0.5mv^{2}$

Since your kinetic energies are the same hence

$0.5m1(v1)^{2}=0.5m2(v2)^{2}$

$m1(v1)^{2}=m2(v2)^{2}$ and making m2 the subject then

$m2=\frac { m1(v1)^{2}}{(v2)^{2}}$

Since v2 is v1+0.28v1=1.28v1

Substituting m1 for 86 Kg

$m2=\frac { 86 Kg(v1)^{2}}{(1.28v1)^{2}}= 52.49023\approx 52.49 Kg$

3 0

3 years ago

A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths 589.0 nm and 589.6 nm. The

DIA [1.3K]

The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

L = 57.88 mm

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

m₁ - m₂ = 2

Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

L = 57.88 mm

Read more at; brainly.com/question/17161594

8 0

1 year ago

How do i do this??????????

Sliva [168]

Power = I^2 x R
Energy = Power x Time

4 0

2 years ago

Ok ok ik ik this is not the answer but I think you need to hear this

leonid [27]

Answer

oh thanksssss i hope u have a great day

Explanation:

ur a really awesome person and i thank u for that

4 0

1 year ago