All categories

2 years ago

If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?

1 answer:

5 0

$\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time, t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0 +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}$

You might be interested in

A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet

aev [14]

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

$E\text{ = }\frac{k|q|}{d^{\prime2}}$

The electric field strength at the midpoint due to the charge q1 will be

$\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint due to the charge q2 will be

$\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the charges will be

$\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

4 0

1 year ago

A skier reaches the end of a 0.9 km track in 1.5 hours, at what speed does it travel?

Wewaii [24]

Speed = Distance/Time

S = D/T

D = 0.9Km, T = 1.5hours,

S = 0.9/1.5

S = 0.6 m/s

hope it helps!

7 0

4 years ago

Read 2 more answers

Please help!!! Explain the type of graph that could be used for the data below.

sdas [7]

Answer:

A table or chart

7 0

3 years ago

Que substâncias deveria escolher para diminuir a basicidade ou para diminuir a acidez de uma solução? *

Gnoma [55]

Answer:

agregar una base

Explanation:

5 0

3 years ago

Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength

Vikentia [17]

Answer:

$\frac{E_{A}}{E_{B}}=4$

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

$E=\frac{F}{q}$ .

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

$F=\frac{kQq}{r^{2}}$ .

By substitution we get that

$E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}$

Now, letting $E_{A}$ be the electric field at point A, letting $E_{B}$ be the electric field at point B, and letting R be the distance from the charge to A:

$E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}$ .

The ration of the electric fields is

$\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4$

This means that at half the distance, the electric field is four times stronger.

4 0

3 years ago