-15 (miles) + 15 (per hour)?

Convert 100°c into Faherenite and kelvin in steps.

wel

Answer:

100°c = 373.15 K

100°C=212°F

Explanation:

Given the 100°C

To convert Celsius to Kelvin, we need the following equation.

°C + 273.15 = K

100°C + 273.15 = K

373.15 = K

Therefore, 100°c = 373.15 K

Celsius To Fahrenhite:

F = 9/5C + 32

=9/5(100)+32

= (180) + 32

= 212°

Therefore,

100°C=212°F

3 0

2 years ago

Your roommate leaves a 120W fan running in your apartment.Over the course of an hour,how much thermal energy does the fan add to

zubka84 [21]

Answer:

$4.32\cdot 10^5 J$

Explanation:

Power is related to energy by the following relationship:

$P=\frac{E}{t}$

where

P is the power used

E is the energy used

t is the time elapsed

In this problem, we know that

- the power of the fan is P = 120 W

- the fan has been running for one hour, which corresponds to a time of

$t = 1 h \cdot (60 min/h)(60 s/min)=3600 s$

So we can re-arrange the previous equation to find E, the energy (in the form of thermal energy) released by the fan:

$E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J$

3 0

3 years ago

Read 2 more answers

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both

Nuetrik [128]

Answer:

Part a)

Width of the slit is

$a = 580 nm$

Part b)

Ratio of intensity is given as

$\frac{I}{I_o} = 0.81$

Explanation:

Part a)

As we know by the formula of diffraction we will have

$a sin\theta = \lambda$

so we have

$\theta = 90$

$\lambda = 580 nm$

so we will have

$a sin90 = 580 nm$

$a = 580 nm$

Part b)

As we know that the intensity in diffraction pattern is given as

$I = I_o (\frac{sin\theta}{\theta})^2$

$\frac{I}{I_o} = (\frac{sin\theta}{\theta})^2$

so for angle 45 degree

$\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2$

$\frac{I}{I_o} = 0.81$

7 0

3 years ago

Can someone help me its really hard to do d is stuff

Tasya [4]

Answer:

direction, speed

means the object is staying still, 0

newtons, N

the sum of all the forces acting on an object

Explanation:

3 0

3 years ago

A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-

musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

$F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}$

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

$F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}$

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0

3 years ago