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4 years ago

13

Velocity describes both speed and direction therefore it is ________ A: same as Acceleration B:A force arrow C: A vector D: A ma

gnitude

1 answer:

Oksi-84 [34.3K]4 years ago

4 0

(A) Acceleration, Remember Speed is the distance covered in a unit of time while acceleration is the rate of change of speed.

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At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago

A sprinter starts from rest and accelerates to her maximum speed of 10.5 m/s in a distance of 11.0 m. (a) What was her accelerat

Anvisha [2.4K]

Answer:

a)a=5.01m/s^2

b)t=11.26s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)

{Vf^{2}-Vo^2}/{2.a} =X(2)

X=Xo+ VoT+0.5at^{2} (3)

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

to solve the question a, we can use the ecuation number 2

Vo=0

Vf=10.5 m/s

x=11m

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{10.5^{2}-0^2}/{2x11} =a

a=5.01m/s^2

to find the time we can use the ecuation number 1

Vf=Vo+a.t

t=(Vf-Vo)/a

t=(10.5-0)/5.01=2.09s

part b

in this case the spees is constant, so the movement is defined by the following ecuation

X=VT

t=x/v

t=96.3/10.5=9.17s

to find the total time we sum the times when the speed is constant and when the acceleration is constan

t=9.17+2.09

t=11.26s

8 0

3 years ago

Which of these letters is the symbol for current in equations A: c B: i C: r D: t

True [87]

Answer: [B]: The letter, "<em /> I " ; for current; in units of "Amps" .
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4 0

3 years ago