After 2.0 s has elapsed on kara's watch, what does kara say the distance is between you and the laser pulse?

A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the

Dimas [21]

Answer:

98N and 147N

Explanation:

We have the following information:

$m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N$

We can find the static fricton force as follow,

$F=\mu_s * N$

Where N is the normal force (mg)

$F=0.3*50*9.8\\F=147N$

Static friction force at 147N is greater than the force applied hence body does not move.

$F=\mu_k N = 0.2*50*9.8= 98N$

3 0

3 years ago

Which statements does the Second Law of Thermodynamics support?

Mazyrski [523]

According to the second law of thermodynamics,
the answer is
4. The entropy of the universe is increasing.

4 0

2 years ago

What is the maximum value of the magnetic field at a distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

3 0

3 years ago

What is the weight of a 42 kg object if the object was on the moon?

sveticcg [70]

(1.6 m/s²)(42 Kg)= 80 N

4 0

3 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

3 0

1 year ago