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Ilia_Sergeevich [38]
3 years ago
10

What is the instantaneous acceleration at t=0?

Physics
2 answers:
ioda3 years ago
8 0

Answer:

0.6 m/sec^2

Explanation:

from the graph we find out the equation of line which will be a function of time and velocity. And differentiating that  we can find acceleration at t=0

two point in in graph are  (0,4) and (10,10)

therefore slope of the graph = 6/10= 3/5

we can find the equation of  line using one point form

v-v_1= m(t-t_1)

putting values we get

v-4= 3/5(t-0)

⇒5v-3t=20

differentiating we get

5\frac{\mathrm{dv} }{\mathrm{d} t}-3=20

\frac{\mathrm{d} v}{\mathrm{d} t}=a=0.6

hence the acceleration = 0.6 m/sec^2

tatyana61 [14]3 years ago
3 0

I have the same physics class , so the answer would be 0.6

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3 years ago
A radio frequency identification application would most likely interface with a (an):_________
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4 0
2 years ago
The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e
Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed (v), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi (1)

Where:

\Delta t - Rotation time, measured in seconds.

R - Radius of the Earth, measured in meters.

\phi - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that \Delta t = 86160\,s, R = 6.371\times 10^{6}\,m and \phi = 37.6819^{\circ}, then the tangential speed at Livermore is:

v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}

v\approx 284.001\,\frac{m}{s}

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0
3 years ago
A ball is thrown vertically up ward with a velocity of 25m/s. a/how fast was it moving after 2sec?​
Shalnov [3]

Answer:

12.5

Explanation:

given

distance=25

time=2sec

required

=speed

solution

=v=s/t

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=12.5m/s

3 0
2 years ago
How much pure gold (24/24) is to be added to 15.0 grams of 14 K gold (14/24) to obtain 18 K gold (18/24)?
Arlecino [84]

Answer:

10 gram gold is added

Explanation:

given data

pure gold = (24/24)

added = 15 grams

gold = 14 K  ( 14/24)

gold = 18 K ( 18/24)

to find out

How much pure gold added

solution

we know here that when we add gold to get 20 K gold or 22K

so we added here 15 gram 14 K

we consider here m is pure gold added

so by composition here

we get

15 (14K)  + m ( 24 K) = ( 15 + m ) (18)    ...................1

solve it and find m

m = 10

so 10 gram gold is added

8 0
3 years ago
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