Helllllllllllllllllp plss

If your speedometer has an uncertainty of 2.5 km/h at a speed of 92 km/h, what is the percent uncertainty

Elis [28]

Answer:

2.7%

Explanation:

Given:

Uncertainty of the speedometer (u)= 2.5km/h

Speed measured at that uncertainty (v) = 92km/h

Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e

p = $\frac{u}{v} * 100$ %

p = $\frac{2.5}{92} * 100$ %

p = 2.7%

Therefore, the percent uncertainty is 2.7%

6 0

3 years ago

A solid sphere rolling without slipping on a horizontal surface. If the translational speed of the sphere is 2.00 m/s, what is i

koban [17]

Answer:

The total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)

Explanation:

The total kinetic energy of a sphere is given by the sum of the rotational kinetic energy and the translational kinetic energy. That is,

$K_{Total} = K_{R} + K_{T}$

The rotational kinetic energy $K_{R}$ is given by

$K_{R} = \frac{1}{2}I\omega^{2}$

Where $I$ is the moment of inertia

and $\omega$ is the angular velocity

The translational kinetic energy $K_{T}$ is given by

$K_{T} = \frac{1}{2}mv^{2}$

Where $m$ is the mass

and $v$ is the translational speed (velocity)

∴ $K_{Total} = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$

But, the moment of inertia $I$ of a sphere is given by

$I = \frac{2}{5}mr^{2}$

Where $m$ is mass

and $r$ is radius

∴ $K_{Total} = \frac{1}{2}\times \frac{2}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}$

$K_{Total} = \frac{1}{5}mr^{2} \omega^{2} + \frac{1}{2}mv^{2}$

Also, $\omega = \frac{v}{r}$

∴ $\omega^{2} = \frac{v^{2} }{r^{2} }$

Then,

$K_{Total} = \frac{1}{5}mr^{2} \times \frac{v^{2} }{r^{2} } + \frac{1}{2}mv^{2}$

$K_{Total} = \frac{1}{5}mv^{2} + \frac{1}{2}mv^{2}$

∴ $K_{Total} = \frac{7}{10}mv^{2}$

From the question, $v = 2.00 m/s$

Then,

$K_{Total} = \frac{7}{10}m(2.00)^{2}$

$K_{Total} = \frac{7}{10}m\times 4.00$

$K_{Total} = 2.8m J$

Hence, the total kinetic energy is 2.8m J. (NOTE: m is mass of the sphere)

8 0

3 years ago

Kawas 200 osebanywhes he remembered he had to return some books they

tia_tia [17]

I think you need to add more.. but I may know where you are leading

Was he 200 m away and made the trip in 200 seconds?

If yes...
2 m/s was his speed and 0 velocity

6 0

3 years ago

Please help (will mark brainliest)

serg [7]

Answer:

if im not mistaken i think its d let me know if correct plz

7 0

3 years ago

Read 2 more answers

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0

3 years ago