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3 years ago

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1 answer:

5 0

1) Swimming at 2 miles per hour
2) Walking at 3 miles per hour 
3) Running at 7 miles per hour 
4) Jogging at 5 miles per hour

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If a zebra weighs 390N what is its mass

stich3 [128]

W=mg
m = W/g = 390N/ 9.8 m/s^2 = 39.8 kg

some people round g to 10 m/s^2 and this will give answer of 39 kg, which is wrong

7 0

3 years ago

A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the

Amanda [17]

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check: V0 = 5 m/s

V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s the speed after 300 sec

3 0

3 years ago

Help me with this please!

alexgriva [62]

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3 0

4 years ago

A horizontal beam of electrons initially moving at 4.0×10^7 m/s is deflected vertically by the vertical electric field between o

givi [52]

Answer:

$1.77\times 10^{-7}\ C/m^2$

0.000439077936334 m

Explanation:

q = Charge of electron = $1.6\times 10^{-19}\ C$

E = Electric field = $2\times 10^{4}\ N/C$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

d = Distance between plates = 2 cm (assumed)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

The beam consists of electrons which means it has negative charge this means the upper plates will be positive and the lower plate will be negative.

The direction is upper to lower lower plate.

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric flux is given by

$\phi=\epsilon_0E\\\Rightarrow \phi=8.85\times 10^{-12}\times 2\times 10^{4}\\\Rightarrow \phi=1.77\times 10^{-7}\ C/m^2$

The charge per unit area on the plates is $1.77\times 10^{-7}\ C/m^2$

Deflection is given by

$s=\dfrac{1}{2}\dfrac{qE}{m}(\dfrac{d}{v})^2\\\Rightarrow s=\dfrac{1}{2}\dfrac{1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}(\dfrac{0.02}{4\times 10^7})^2\\\Rightarrow s=0.000439077936334\ m$

The deflection is 0.000439077936334 m

7 0

3 years ago

What kind of model is shown below?

Kay [80]

Still there ain’t no picture

7 0

3 years ago

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