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2 years ago

Identify the wave with the shortest wave length.

Physics

2 answers:

Valentin [98]2 years ago

7 0

Answer:

d) gamma

Explanation:

hope i helped

lubasha [3.4K]2 years ago

6 0

Answer:

2nd one

mark me brainliest

Explanation:

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If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:

Orlov [11]

I could be wrong but I believe it’s 1/2

6 0

2 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

2 years ago

If you are six feet tall how far back from a 3 foot mirror do you have to stand in order to see yourself completely?

OverLord2011 [107]

Answer:

you would have to stand 6 ft back

Explanation:

7 0

2 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

a ship A , steaming in a direction 030° with a steady 12 km/h, sight a ship B. The velocity of B relative to A is 10km/h in a di

pickupchik [31]

Answer: 11 km/h at 339° compass

Explanation:

A sees B moving south at 0 km/h

A is moving north at 12cos30 = 10.392 km/h

Therefore B must be moving north at 10.392 k/h

A is moving east at 12sin30 = 6 km/h

B appears to be moving west at 10 km/h

Therefore B must be moving west at 10 - 6 = 4 km/h

B is moving v = √(4² + 10.392²) = 11.135... 11 km/h

θ = arctan( -4 / 10.392) = -21.05 = 339°

8 0

2 years ago