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3 years ago

7

What is the best way to ensure that the results of your experiment are accurate?

1 answer:

Ainat [17]3 years ago

6 0

Re do the experiment 3 times over again and if you get the same results then it should be correct

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When you trace the outline of your palm how do you find its area

oksian1 [2.3K]

Answer:

Explanation:

if its squares count the squares els messure it i think

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3 years ago

Which of these activities increases the amount of carbon in the atmosphere? A. animals eating plants B. burning of fossil fuels

tester [92]

Answer:

B. Burning of fossil fuels

8 0

3 years ago

An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an

Luba_88 [7]

Answer:

F = 4399 KN

Explanation:

given,

mass of automobile = 890 kg

initial speed = 48 km/h

= 48 × 0.278 = 13.34 m/s

using equation of motion

v² = u² + 2 a × s

0 = 13.34² - 2 a ×0.018

$a = \dfrac{13.34^2}{0.036}$

a = 4943.21 m/s²

F = m × a

F = 890 × 4943.21

F = 4399456.9 N

F = 4399 KN

hence, the Net force is F = 4399 KN

4 0

3 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

6 0

3 years ago

All organisms require a set of instructions that specify its traits. The instructions that are responsible for all the inherited

garri49 [273]

The genes that are inside the chromosomes of the cell.

5 0

3 years ago

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