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4 years ago

Which of the following is an example of velocity?

Physics

2 answers:

mina [271]4 years ago

8 0

Answer:

C. 10 m/s to the right.

Explanation:

I took the test

Otrada [13]4 years ago

5 0

Answer:

10 m/s to the right

Explanation:

Velocity is a vector quantity it has magnitude as well as direction.

10 m/s represent the speed of the object not velocity.

10 is a constant so it is not an example of the velocity.

10 m to the right is an example of displacement of the object not velocity.

10 m/s to right represent magnitude as well as direction so, its an example of the velocity.

Thus correct answer is 10 m/s to right.

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Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau

Tamiku [17]

Answer:

$7.0\cdot 10^{-13}C$

Explanation:

The magnitude of the electrostatic force between two charged objects is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

$r=0.070 cm =7\cdot 10^{-4} m$ is the distance between the charges

$q_1=q_2=q$ since the charges are identical

$F=9.0\cdot 10^{-9}N$ is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

$F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C$

8 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

A jet plane is cruising at 310 m/s when suddenly the pilot turns the engines up to full throttle. after traveling 4.0 km , the j

Inessa [10]

For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad
where:
Vf = 400m/s
Vi = 300m/s
a = ?
d = 4.0km
= 4000m

400^2 = 300^2 + 2a4000
a = [ 160000 - 90000 ] / 8000
a = 8.75m/s^2
rounding it off to 2 significant figures, will give us 8.8 m/s^2.

4 0

4 years ago

An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i

Diano4ka-milaya [45]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

First of all, we can use the lens equation to find the location of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$