Stars emit different wavelengths of visible light due to their different

Answer<br> It's too short. Write at least 20 characters to get a great answer.

Roman55 [17]

Answer: I think the first option is the best option because water isn't a heat conductor and all the other options are not very accurate.

Explanation:

8 0

3 years ago

What is the constant snell’s law

Marat540 [252]

Snell's law is defined as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media”.

3 0

2 years ago

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble s

Verdich [7]

There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

Answer:

w=50 rpm

Explanation:

Given data

The mass turntable M=2kg

Diameter of the turntable d=20 cm=0.2 m

Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s

Two blocks Mass m=500 g=0.5 kg

To find

Turntable angular velocity

Solution

We can find the angular velocity of the turntable as follow

Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

$I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}$

where I is moment of inertia

$w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm$

7 0

3 years ago

If the initial upward speed of the ball in Activity 7 C.2 is 10m/s, and the ball is releasedat a height of 1.5 m above the oor,

Ulleksa [173]

Answer:

6.5 m above the floor and 5 m above Christine's hand when it reaches the maximum height.

Explanation:

Let g = 10 m/s2 be the gravitational deceleration that affects the ball vertical motion so it comes to the maximum height at 0 speed. We can use the following equation of motion to find out the distance traveled by the ball from where it's thrown:

$v^2 - v_0^2 = 2g\Delta s$

where v = 0 m/s is the final velocity of the ball when it reaches maximum level, $v_0$ = 10m/s is the initial velocity of the ball when it starts, g = -10 m/s2 is the deceleration, and $\Delta s$ is the distance traveled, which we care looking for:

$0^2 - 10^2 = -2*(-10)\Delta s$

$\Delta s = 100 / (2 * 10) = 5 m$

So the ball is 5 m above Christine' hands when it reaches maximum height, and since the hand is 1.5 m above the floor, the ball is 5 + 1.5 = 6.5 m above the floor when it reaches maximum height.

5 0

3 years ago

Read 2 more answers

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago