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3 years ago

Can somebody help me with this

Physics

1 answer:

Ivanshal [37]3 years ago

6 0

this is simple the word your looking for is 8 letters long and the definition is key to the answer

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Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins

irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

But dy/dt = velocity.

If dy/dt = 0, then

60t - 15 = 0

60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

4 0

3 years ago

Which type radiation can be observed well from Earth's surface?

umka2103 [35]

Answer:

Eletromagnetic radiation which is also known as visible light.

Explanation:

4 0

3 years ago

Read 2 more answers

Convert 1,265,341 mm to m.

Thepotemich [5.8K]

The answer in Meters is going to to 1265.341

7 0

3 years ago

Read 2 more answers

A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. (a) What is the acceleration exp

givi [52]

Yes the plot dose make it with out baking out

7 0

3 years ago

The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall nea

djverab [1.8K]

Answer:

Acceleration, $a=9.39\ m/s^2$

Explanation:

Given that,

Mass of the planet Krypton, $m=8.8\times 10^{23}\ kg$

Radius of the planet Krypton, $r=2.5\times 10^{6}\ m$

Value of gravitational constant, $G=6.6726\times 10^{-11}\ Nm^2/kg^2$

To find,

The acceleration of an object in free fall near the surface of Krypton.

Solution,

The relation for the acceleration of the object is given by the below formula as :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}$

$a=9.39\ m/s^2$

So, the value of acceleration of an object in free fall near the surface of Krypton is $9.39\ m/s^2$

5 0

3 years ago