Ask question

Ask question

All categories

3 years ago

11

In the International System of Units, the basic unit for measuring mass is what

1 answer:

nadezda [96]3 years ago

3 0

Answer:

Kilograms (kg)

Explanation:

You might be interested in

(i) a force of 35.0 n is required to start a 6.0-kg box moving across a horizontal concrete floor, (a) what is the coefficient o

Serjik [45]

a. The force applied would be equal to the frictional force.

F = us Fn

where, F = applied force = 35 N, us = coeff of static friction, Fn = normal force = weight

35 N = us * (6 kg * 9.81 m/s^2)

us = 0.595

b. The force applied would now be the sum of the frictional force and force due to acceleration

F = uk Fn + m a

where, uk = coeff of kinetic friction

35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)

uk = 0.533

4 0

3 years ago

Stars are classified mainly according to their distance from Earth.

Zielflug [23.3K]

The answer is b) false. Stars are mainly classified by their spectra (the elements they absorb) and their temperature.

4 0

3 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

5 0

3 years ago

I would like to know the answers to both 10 & 11

oksian1 [2.3K]

Answer:

picture is not showing up sorry

Explanation:

7 0

4 years ago

HELP PLEASE!!!! 20 POINTS! If we increase the distance traveled over the same period of time, this will (2 points) decrease the

Eddi Din [679]

Increase the speed of it

7 0

3 years ago