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elena-14-01-66 [18.8K]

3 years ago

11

gabriel and arely have performed an experiment and are not sure their results are valid. what should they do to check their resu

lts

2 answers:

zubka84 [21]3 years ago

7 0

Do the math and then compare your results to theirs.

Arisa [49]3 years ago

6 0

Yeah, or they can compare the results to other people

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The half-life of Cs-137 is 30.2 years. If the initial mass of the sample is 1.00 kg, how much will remain after 151 years?

Xelga [282]

Answer:

i found something on a website that will help you

Explanation:

do you want me to send you it???

3 0

3 years ago

Read 2 more answers

s2008m [1.1K]

Answer:

55N

Explanation:

Using Newton's second law of motion:

F=ma

Force=mass × acceleration

F=25×2.2

F=55N

So 55 Newtons are needed

8 0

3 years ago

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

vagabundo [1.1K]

Answer:

R_total = 14.57 Ω , V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance

$R_{eq}$ = ∑ $R_{i}$

* For resistance in parallel

1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

R_B = Rb + R4

R_B = 2 + 18

R_B = 20 Ω

Branch C

R_C5 = Rc + R5

R_C5 = 3 + 12

R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

1 / R_BC = 1 / R_B + 1 / R_C5

1 / R_BC = 1/20 + 1/15 = 0.116666

R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

R_total = R_A + R_BC

R_total = 6 + 8.57

R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

10 = V_a + V_BC

10 = i R_a + i R_BC = i (R_a + R_BC)

i = 10 / (R_a + R_BC)

i = 10 / (14.57)

i = 0.6863 A

The current in the series circuit is constant

V_BC = i R_BC

V_BC = 0.6863 8.57

V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

Branch C

V_BC = i R_C5

i = V_BC / R_C5

i = 5.8819 / 15

i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

V_C = i R_C

V_C = 0.39213 3

V_C = 1.176 V

3 0

3 years ago

Why is Energy not created nor destroyed?<br> (Need the answer ASAP!!)

eduard

Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the surroundings outside the system. In other words, energy cannot be created or destroyed.

4 0

3 years ago

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago