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3 years ago

How will you change your picture

2 answers:

8 0

If I remember right you go to edit profile click on photo and then drag and drop or copy and past something along those lines.

Grace [21]3 years ago

6 0

Click on your picture and your mobile photos will become to appear. Select the picture you want or like by double tap....

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Boy or girl and age you can still comment if its answered

Ksenya-84 [330]

I am a girl and i have a virgina and im 13 tho so dont touch me lolz mark brainly

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2 years ago

A car stopped at a red light, not moving?

timofeeve [1]

I think it's a) 1st Newton's law... so sorry if it's wrong...

5 0

3 years ago

A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

2 years ago

Two objects have the same momentum. Do the velocities of these objects necessarily have (a) the same directions and (b) the same

NeX [460]

Answer:

(a) They must have same direction

(b) It is not necessary for them to have same magnitudes

Explanation:

(a)

Momentum is a vector quantity. It is the product of mass (scalar) and velocity (vector). Thus, if the direction of velocity is changed, then as a result the direction of momentum will also change or its magnitude or component in the same direction will change. Hence, for the two objects to have same momentum, the directions of their velocities must also be the same.

(b)

Since, the momentum is product of velocity and mass. It is possible that two bodies of different masses with different velocities might have same momentum, provided the direction of their velocities is same.

For example, take a body of mass 4 kg moving with speed 5 m/s. It will have a momentum of 20 N.s. Now, consider another body of mass 2 kg, moving with speed 10 m/s. It will also have a momentum of 20 N.s.

Thus, it is not necessary for two objects to have same magnitude of velocity to have same momentum.

3 0

3 years ago

A wire that is 0.36 meters long moves perpendicularly through a magnetic field at a speed of 0.21 meters/second. The induced emf

dmitriy555 [2]

0.034 newtons/amp·meter

8 0

3 years ago

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