Last month Dr. Hruda's electric bill showed that she used 249 kW-hr. How many Joules (kJ) did Dr. Hruda use?

Falling objects drop with an average acceleration of 9.8 m/sec/sec, or 9.8 m /s2. If an object falls from the top of a tall buil

Lena [83]

Use the velocity expression for uniform acceleration and solve for t: v = v0 + at. v0 is zero since the object is at rest. 49 m/s = a(t), and solve for t. t = 49 m/s / 9.8 = 5 seconds.

5 0

3 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity

umka21 [38]

Answer: $29.85\ W/m^2$

Explanation:

Given

Power $P=60\ W$

Distance from the light source $r=0.4\ m$

Intensity is given by

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

Inserting values

$\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2$

3 0

3 years ago

Read 2 more answers

A condition that affects the ability to sleep or the quality of sleep is referred to as a __________.

Anit [1.1K]

The answer is B) Sleep disorder. Notice that is says, "A condition that affects the ability to sleep or quaility of sleep." Meaning how much sleep you get. Also known as sleep disorder. Answer is B. Hope that helps :)

7 0

3 years ago

Read 2 more answers

A point charge is placed at the center of a spherical Gaussian surface. The electricflux ΦEischangedif(a) a second point charge

Simora [160]

Answer:

(b) the point charge is moved outside the sphere

Explanation:

Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.

$\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.

If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.

Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.

7 0

3 years ago

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$

The magnitude of the displacement vector = $\sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m$

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0

3 years ago