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sasho [114]
3 years ago
9

9. A car accelerates from 3.5 m/s to 17 m/s in 4.5 seconds. Find the acceleration of the car in m/s2.

Physics
1 answer:
Citrus2011 [14]3 years ago
8 0
The answer is -3 m/s2
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Two​ vehicles, a car and a​ truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe
olya-2409 [2.1K]

The car heads east at an average speed of 50 miles per​ hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per​ hour from the intersection point towards South.

The distance of car from the intersection point after t hours is 50t.

The distance of truck from the intersection point after t hours is 60t.

Since these distances are perpendicular to each other, distance apart d​ (in miles) at the end of t hours is

d=\sqrt{(50t)^2+(60t)^2} \\ d=10\sqrt{61} t\\ d=78.1t

Thus the distance apart is d=78.1t \;miles

5 0
3 years ago
Most of the substances around you are _______
monitta

Answer:

gravity

Explanation:

5 0
3 years ago
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A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of
levacccp [35]

\underline {\huge \boxed{ \sf \color{skyblue}Answer :  }}

<u>Given :</u>

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{D} = 5mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{DSS} = 10mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS(off)} = -6V

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS} =   {?}

\:  \:  \:

<u>Let's Slove :</u><u> </u>

  • \tt \large  I_{D} = I_{(DSS)}  (1 -   \frac {V_{GS}}{V_{GS(off)}} )^{2}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times  V_{GS(off)}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times  { - 6}

\:  \:

  • \underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}
3 0
1 year ago
If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun
inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

s=-8t^2+32t...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

\dfrac{ds}{dt}=0

\dfrac{d(-8t^2+32t)}{dt}=0

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

s=-8(2)^2+32(2)

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

6 0
3 years ago
An AC generator consists of eight turns of wire, each of area 0.0775 m2 , and total resistance of 8.53 Ω. The loop rotates in th
Bad White [126]

Answer:

44.08 Volt

Explanation:

N = 8, A = 0.0775 m^2, R = 8.53 ohm, B = 0.222 T, f = 51 Hz

e0 = N B A w

e0 = 8 x 0.222 x 0.0775 x 2 x 3.14 x 51

e0 = 44.08 Volt

3 0
3 years ago
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