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kondor19780726 [428]

2 years ago

6

Why is velocity a vector quantity

2 answers:

Firlakuza [10]2 years ago

6 0

Answer:

velocity is called vector quantity because it has both magnitude and direction.

Jobisdone [24]2 years ago

4 0

Velocity as a Vector Quantity
Because the person always returns to the original position, the motion would never result in a change in position. Since velocity is defined as the rate at which the position changes, this motion results in zero velocity.

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If 41.5 mol of an ideal gas occupies 86.5 L at 27.00 °C, what is the pressure of the gas?

Dmitry [639]

PV=nRT
(P)(86.5)=(41.5)(.08206)(300.15)
(P)(86.5)=(1022.157824)
P=11.81685345 atm

7 0

2 years ago

Read 2 more answers

The elements with the largest atomic radii are found in the

Ulleksa [173]

Elements with the largest atomic radius are found in the lower left hand of the periodic table.

7 0

2 years ago

When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b

Nesterboy [21]

Answer:

Explanation:

Given

Displacement is $\frac{1}{3}$ of Amplitude

i.e. $x=\frac{A}{3}$ , where A is maximum amplitude

Potential Energy is given by

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}k(\frac{A}{3})^2$

$U=\frac{1}{18}kA^2$

Total Energy of SHM is given by

$T.E.=\frac{1}{2}kA^2$

Total Energy=kinetic Energy+Potential Energy

$K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2$

$K.E.=\frac{8}{18}kA^2$

Potential Energy is $\frac{1}{8}$ th of Total Energy

Kinetic Energy is $\frac{8}{9}$ of Total Energy

(c)Kinetic Energy is $0.5\times \frac{1}{2}kA^2$

$P.E.=\frac{1}{4}kA^2$

$\frac{1}{2}kx^2=\frac{1}{4}kA^2$

$x=\frac{A}{\sqrt{2}}$

7 0

2 years ago

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The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

2 years ago

Can someone please help?

Diano4ka-milaya [45]

I think you go to my school!

7 0

2 years ago