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Salsk061 [2.6K]
3 years ago
7

What is the average speed in 4 seconds going 8 meters

Physics
2 answers:
gulaghasi [49]3 years ago
6 0

Answer:

Explanation:

speed is define as rate of change of distance or displacement

v=s/t

s=8 m

t=4s

v=8/4

v=2 m/s

Ksenya-84 [330]3 years ago
4 0
The formula for speed is distance over time or s = d/t
Distance = 8m or meters
Time: 4s or seconds
So s = 8/4
s = 2m/s or meters per second
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All of the following would be useful in trying to obtain procedural information to replicate an experiment previously published
Lera25 [3.4K]

Answer:

I am pretty sure that the answer would be a peer reviewed article.

Explanation:

I saw this because, an encyclopedia, published scientific journal, and a lab journal used in the original experiment, are all reliable sources of information.

7 0
3 years ago
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We have a long wire with a circular cross section and radius a = 2.40 cm. The current density in this wire is uniform, with a to
Alika [10]

Answer:

The magnitude of the magnetic field is 7.49x10⁻⁶T

Explanation:

The magnetic field is:

B=(\frac{\mu i}{2\pi R^{2} } )r

Where

i = current = 3 A

R = radius = 2.4 cm = 0.024 m

r = distance = 0.72 cm = 7.2x10⁻³m

Replacing:

B=(\frac{4\pi x10^{-7} *3}{2\pi *0.024^{2} } )*7.2x10^{-3}= 7.49x10^{-6} T

7 0
3 years ago
The relative density of oxygen and carbon dioxide are 16, 12 respectively. If 25cm3 of carbon dioxide effuse out in 75 sec what
Mrac [35]

Answer:

32 cm³

Explanation:

The given gas data are;

The relative density of oxygen = 16

The relative density of carbon dioxide = 12

The time it takes 25 cm³ of carbon dioxide to effuse out = 75 seconds'

The duration of effusion of the oxygen = 96 seconds

The rate of effusion of carbon dioxide, R1 = 25 cm³/(75 sec) = (1/3) cm³/sec

According to Graham's law of diffusion and effusion of a gas, we have;

\dfrac{Rate \ of \ effudion \ of \ gas \ 1}{Rate \ of \ effudion \ of \ gas \ 2} =\dfrac{The \ relative \ density \ of \ gas \ 2}{The \ relative \ density \ of \ gas \ 1}

Therefore, we have;

\dfrac{Rate \ of \ effudion \ of \ oxygen}{(1/3)} =\dfrac{12}{16}

The \ rate \ of \ effudion \ of \ oxygen}=\dfrac{12}{16} \times \left(\dfrac{1}{3 } \ cm^3/sec\right ) = \dfrac{1}{4} \ cm^3/sec

The volume of effusion = The rate of effusion × Time

The volume of the oxygen that will effuse in 96 seconds is given as follows;

The rate of effusion of a gas × Time

V = The rate of effusion of oxygen × Time = (1/3) cm³/sec × 96 sec = 32 cm³

The volume of oxygen that will effuse in 96 seconds, V = 32 cm³.

8 0
3 years ago
A rabbit trying to escape a fox runs north for 8.0 m, darts northwest for 1.0 m, then drops 1.0 m down a hole into its burrow. W
kifflom [539]

Answer:

9.61 m

Explanation:

d1 = 8 m north

d2 = 1 m north west

d3 = 1 m vertically down

Write the displacements in vector form

\overrightarrow{d_{1}}=8\widehat{j}m

\overrightarrow{d_{2}}=1 \left ( -Cos45\widehat{i}+Sin45\widehat{j} \right )m

\overrightarrow{d_{2}}=\left ( -0.707\widehat{i}+0.707\widehat{j} \right )m

\overrightarrow{d_{3}}=-1\widehat{k}m

The resultant displacement is given by

\overrightarrow{d}=\overrightarrow{d_{1}}+ \overrightarrow{d_{2}} +\overrightarrow{d_{3}}

\overrightarrow{d}=-0.707\widehat{i}+ 8.707\widehat{j} - 4\widehat{k}m

The magnitude of displacement is given by

d =

\sqrt{\left ( -0.707 \right )^{2}+\left ( 8.707 \right )^{2}+\left ( -4 \right )^{2}}

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4 0
3 years ago
What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is
ss7ja [257]

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R²   ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

<u>g' = 1.09 m/s²</u>

3 0
3 years ago
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