All categories

3 years ago

Describe the steps you use during an experiment

Physics

2 answers:

Nezavi [6.7K]3 years ago

5 0

Gather your materials
Do the experiment
Collect the data

DochEvi [55]3 years ago

4 0

<span>come up with a question you want answered, create a hypothesis, perform the experiment to test your hypothesis. obtain your data and analysis. write down your results and compare them to your hypothesis. then come up with a conclusion whether your hypothesis was right or wrong.</span>

You might be interested in

Vector of magnitude 15 is added to a vector of magnitude 25. The magnitude of this sum

loris [4]

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

$R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45$

The minimum sum of the resultant vector,

$R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10$

So, the magnitude of this sum either 45 or -10.

6 0

2 years ago

Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi

Vedmedyk [2.9K]

Answer:

$v = 1.98\ m/s$

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m u₄ = 4.5 m/s

m₅ = m u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

$v = \dfrac{9.9}{5}$

$v = 1.98\ m/s$

5 0

2 years ago

Which formation is one feature of karst topography?

steposvetlana [31]

Karst is a topography formed from the dissolution of soluble rocks such as limestone, dolomite, and gypsum. It is characterized by underground drainage systems with sinkholes and caves. It has also been documented for more weathering- resistant rocks, such as quartzite, given the right conditions.

8 0

2 years ago

Read 2 more answers

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago