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3 years ago

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Please help ASAP will mark brainliest How do we find the amount of uncertainty we should expect in our measurement of the period

? Come up with and explain a method to estimate the uncertainty.

1 answer:

kicyunya [14]3 years ago

4 0

Answer:

The relative uncertainty gives the uncertainty as a percentage of the original value. Work this out with: Relative uncertainty = (absolute uncertainty ÷ best estimate) × 100%. So in the example above: Relative uncertainty = (0.2 cm ÷ 3.4 cm) × 100% = 5.9%. The value can therefore be quoted as 3.4 cm ± 5.9%.

Explanation:

hope it helps :)

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Describe the difference between a conductor and an insulator in terms of thermal energy transfer and provide an example of each.

Nikitich [7]

The difference between conductors and insulators is because of electronic structure of atoms and molecules.

Anywhere we have free electrons, we have a hard conductor where free electrons help in conduction.

In electrolytes, charged Ions do conduction.

Dielectrics (or insulators) don't have free electrons current.

Free electrons mean, the electrons that are not related with a specific atom, they are allowed to move nearby the crystal lattice.

4 0

3 years ago

A disk with a uniform positive surface charge density lies in the x-y plane, centered on the origin. The disk contains 2.5 x 10-

allochka39001 [22]

Answer:

E=3 x 10^4 N/c

Explanation:

The electric field strength can be found out disk with a uniform positive surface charge density by

$E= (\sigma/\2epsilon_o)(1-z/ \sqrt(z^2+r^2))$

σ= charge density

r= radius of the disk

z= position in which we have to find electric field = 15 cm

ε_0= constant ( vacuum permitivity)

putting values we get

$E= \frac{2.5\times10^{-6}}{2\times2.5\times10^{-6}}(1-\frac{0.15}{\sqrt{0.15^2+0.075^2} })$

solving we get

E=30000 N/c

E=3 x 10^4 N/c

6 0

3 years ago

A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one

alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R= $13\Omega$

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper wire=Sum of resistance of two piece of wires

$r+7r=13$

$8r=13$

$r=\frac{13}{8}$ ohm

Resistance of long piece of wire= $7\times\frac{13}{8}=\frac{91}{8}\Omega$

Resistance of short piece of wire = $\frac{13}{8}\Omega$

Resistivity of wire and cross section area of wire remains same .

Let L be the total length of wire and L' be the length of short piece of wire.

We know that

$R=\frac{\rho L}{A}$ = $\frac{\rho}{A}L=KL$

$\frac{R}{L}=K$

Where K= $\frac{\rho}{A}$ =Constant

Using the formula

$\frac{13}{L}=\frac{\frac{13}{8}}{L'}$

$\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}$

$L'=\frac{L}{8}$

Length of short piece of wire=L'= $\frac{L}{8}$

Length of long piece of wire= $L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L$

% of length of short piece of wire= $\frac{\frac{L}{8}}{L}\times 100=12.5%$ %

The resistance of the short piece= $\frac{13}{8}\Omega$

The resistance of the long piece= $\frac{91}{8}\Omega$

8 0

3 years ago

Stopping distance of vehicles When brakes are applied to a moving vehicle, the distance it travels before stopping is called sto

DochEvi [55]

Answer:

Stopping distance = 40m

Explanation:

Given the following :

Initial speed of vehicle before applying brakes = 72km/hr

Converting km/hr to m/s:

72km/hr = [(72 * 1000)m] / (60 * 60)

72km/hr = 72,000m / 3600s

72km/hr = 20m/s

Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2

From the 3rd equation of motion:

v^2 = u^2 + 2as

Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance

Final velocity when the car stops will be 0

Therefore ;

v^2 = u^2 + 2as

0 = 20^2 + 2(-5)(s)

0 = 400 - 10s

10s = 400

s = 400/10

s = 40m

Therefore, the stopping distance of the car = 40 meters

7 0

3 years ago

Two ice skaters collide on the ice. A 39.6-kg skater moving South at 6.21 m/s collides with a 52.1-kg skater moving East at 4.33

Stells [14]

Answer:

V = 3.6385 m/s

θ = 47.46 degrees

Explanation:

the important data in the question is:

Skater 1:

$M_1$ = 39.6 kg

direction: south (axis y)

$V_{1iy}$ = 6.21 m/s

Skater 2:

$M_2$ = 52.1 kg

direction: east (axis x)

$V_{2ix}$ = 4.33 m/s

Now using the law of the conservation of linear momentum ( $P_i = P_f$ and knowing that the collision is inelastic we can do the next equations:

$M_{1}V_{1ix}+M_2V_{2ix} = V_{sx}(M_1+M_2)$ (eq. 1)

$M_{1}V_{1iy}+M_2V_{2iy} = V_{sy}(M_1+M_2)$ (eq. 2)

Where $V_{sx}$ and $V_{sy}$ is the velocity of the sistem in x and y after the collision.

Note: the conservation of the linear momentum have to be make once by each axis.

Now, in the (eq. 1) the skater 1 don't have velocity in the axis x, so we can replace $V_{1ix}$ by 0 in the equation and get:

$M_2V_{2ix} = V_{sx}(M_1+M_2)$ (eq. 1)

also, in the (eq. 2) the skater 2 don't have velocity in the axis y, so we can replace $V_{2iy}$ by 0 in the equation and get:

$M_{1}V_{1iy} = V_{sy}(M_1+M_2)$ (eq. 2)

Now, we just replace the data in both equations:

$(52.1)(4.33) = V_{sx}(39.6+52.1)$ (eq. 1)

$(39.6)(6.21) = V_{sy}(39.6+52.1)$ (eq. 2)

solving for $V_{sx]$ and $V_{sy}$ we have:

$V_{sx]$ = 2.46 m/s

$V_{sy]$ = 2.681 m/s

using the pythagoras theorem we can find the magnitude of the velocity as:

V = $\sqrt{2.46^2+2.681^2}$

V = 3.6385 m/s

For find the direction we just need to do this;

θ = $tan^{-1}(\frac{2.681}{1.46})$

θ = 47.46 degrees

3 0

3 years ago