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3 years ago

13

What are the relationships between position, velocity, and acceleration as it relates to time?

2 answers:

Ber [7]3 years ago

5 0

Velocity = distance /time
acceleration = velocity / time

Lostsunrise [7]3 years ago

5 0

Your velocity is the rate at which your position changes over time, and your acceleration is the rate at which your velocity changes over time :)

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Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

2 years ago

A bag of rocks has a mass of 16.4 kg what is it weight here on the earth

Alex73 [517]

Answer
160,72N
Explanation
W=mg
=(16.4)(9.8)

6 0

1 year ago

ivann1987 [24]

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3 years ago

Read 2 more answers

A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene

Yuri [45]

Explanation:

The given data is as follows.

$U_{a} = 5.4 \times 10^{-8} J$

$W_{/text{a to b}} = -1.9 \times 10^{-8} J$

Electric potential energy ( $U_{b}$ ) = ?

Formula to calculate electric potential energy is as follows.

$U_{b} = U_{a} - W_{/text{a to b}}$

= $5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)$

= $7.3 \times 10^{-8} J$

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is $7.3 \times 10^{-8} J$ .

6 0

3 years ago

A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0

3 years ago