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Sindrei [870]
3 years ago
7

Consider this line graph that represents speed in meters/second. The X-axis is labelled in seconds; the Y-axis is labelled in me

ters. Imagine that the moving object stops after 2.5 seconds and stays stopped for 0.5 seconds. What would the line for that time interval look like on this graph?
Physics
2 answers:
Mrac [35]3 years ago
8 0
If the x-axis represent time and the y-axis represents distance,
then the SPEED in meters/second is the SLOPE of the line
on the graph.

The object is moving, so the line is either rising or falling, but
it's definitely not staying horizontal, because the distance is
changing.

The object stops moving at 2.5 seconds, and stays stopped
for 0.5 second.  So right at the 2.5 sec point, the line stops
sloping, and becomes horizontal until 3.0 sec.

We don't know what happens then.  If the object stays still
for longer, then the line stays horizontal until the object starts
moving again.  Then the line starts sloping again.
Ivanshal [37]3 years ago
8 0

To represent no movement of distance/time graph, the line would have no slope. In this case, the line would be straight and horizontal at Y equals 14, to {3, 14}.

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0.4 m/s

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4 0
3 years ago
A solid sphere of radius 30cm is uniformly charged to 100nC. a) What is the volume charge density of the sphere? b) What is the
g100num [7]

Answer

given,

total charge Q = 100 n C

                        = 100 × 10⁻⁹ C

radius of the solid sphere = 30 cm

                                           = 0.3 m

Volume of sphere = \dfrac{4}{3}\pi r^3

                              = \dfrac{4}{3}\pi\times 0.3^3

                              =0.113 m³

a) volume charge density

\rho = \dfrac{10^{-7}}{0.133}

         ρ  = 8.85 × 10⁻⁷ C/m³

b) at r = 10 cm = 0.1 m

charge in the sphere at radius

Q = \dfrac{4}{3}\pi\times 0.1^3\time \rho

   = 3.7037 \times 10^{-9}C[/tex]

Field strength

E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{3.7037 \times 10^{-9}}{4\pi \times 8.85\times 10^{-12}\times 0.1^2}

      = 3.33 \times 10^3 N/C

at r = 20 cm = 0.2 m

Q = \dfrac{4}{3}\pi\times r^3\time \rho

E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{ \dfrac{4}{3}\pi\times r^3\time \rho}{4\pi \epsilon_0 r^2}

E_1 = \dfrac{\rho}{3 \epsilon_0}

E_1 = \dfrac{0.2\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}

      = 6.66 \times 10^3 N/C

at r = 30 cm

E_1 = \dfrac{\rho}{3 \epsilon_0}

E_1 = \dfrac{0.3\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}

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6 0
4 years ago
The acceleration due to gravity at the surface of a planet depends on the planet's mass and size; therefore other planets will h
Ahat [919]

Answer:1.88 m/s^2

Explanation:

Given

initial horizontal speed of =6.55  m/s

height of rock=1.4 m

Horizontal distance=8 m

Time to travel 8 m

8=6.55\times t

t=\frac{8}{6.55}=1.22 s

Time to cover 1.4 m

h=\frac{at^2}{2}

1.4=\frac{g'1.22^2}{2}

g'=1.88 m/s^2

6 0
4 years ago
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