Static friction is the friction between an object at rest and the surface on which it is resting on.

Kinetic friction is the friction between a moving object and the surface on which it moves.

Hopefully you understood and this helps :)

3 0

4 years ago

An object with a mass of 5.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the acceleration? (20 poin

Readme [11.4K]

Answer:

40N

Explanation:

F=ma

F= 5 × 8 = 40N

.......

4 0

3 years ago

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

$C=\frac{Q_0}{V_0}$

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

$C=4.4\mu F = 4.4\cdot 10^{-6}F$

$V_0 = 5.1 V$

Solving for $Q_0$ ,

$Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C$

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

where

$Q_0 = 22.4 \mu C$

t is the time

$R=2.0 \Omega$ is the resistance

$C=4.4\mu F$ is the capacitance

Therefore, at time $t=1 \mu s$ , we have:

$Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C$

3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

However, here the time to consider is

$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

4 0

3 years ago

The drawing shows two long, thin wires that carry currents in the positive z direction. Both wires are parallel to the z axis. T

erica [24]

Answer:

The magnitude of the magnetic field at the origin is $2.56\times 10^{-6}\ T$ .

Explanation:

Given :

50-A wire is in the x-z plane and is 5 m from the z axis.

Also , 40-A wire is in the y-z plane and is 4 m from the z axis.

Now , since both the wire are perpendicular to each other .

Therefore , magnetic field are also perpendicular to each other .

Magnetic field at origin due to wire 1 is :

$B_1=\dfrac{\mu_oI_1}{2\pi R_1}\\\\B_1=\dfrac{(50)\mu_o}{2\pi( 5)}\\\\B_1=\dfrac{5\mu_o}{\pi}$

Magnetic field at origin due to wire 2 is :

$B_2=\dfrac{\mu_oI_2}{2\pi R_2}\\\\B_2=\dfrac{(40)\mu_o}{2\pi( 4)}\\\\B_2=\dfrac{4\mu_o}{\pi}$

Now , therefore net magnetic field is :

$B=\sqrt{B_1^2+B_2^2}\\\\B=\sqrt{(\dfrac{5\mu_o}{\pi})^2+(\dfrac{4\mu_o}{\pi})^2}\\\\B=\dfrac{\sqrt{41}\mu_o}{\pi}$

Putting value of $\mu_o=4\pi \times 10^{-7}\ H/m$

We get ,

$B=\sqrt{41}\times 4\times 10^{-7}\\B=2.56\times 10^{-6}\ T$

Therefore, the magnitude of the magnetic field at the origin is $2.56\times 10^{-6}\ T$ .

5 0

3 years ago