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2 years ago

In which material does the light wave have the larger wavelength?

Physics

1 answer:

pochemuha2 years ago

8 0

Answer: in the smaller/ thinner pieces.

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A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

3 years ago

A student measures the speed of sound by echo destiny classes hands and then measures the time to hear the echo his distance to

777dan777 [17]

Explanation:

∆x=300 m×2

∆t=1.5 s

v=∆x/∆t → v=2×300/1.5 = 400 m/s

6 0

3 years ago

In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (

omeli [17]

Answer:

58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

V=58

7 0

3 years ago

How much force is required to accelerate a 5 kg mass at 20 m/s^2

Studentka2010 [4]

Hello!

$\large\boxed{F = 100N}$

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:

F = 5 · 20

F = 100N

6 0

3 years ago

a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':

Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

$v=\sqrt{u^{2}+2as}$

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

$v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s$

Also, v=u+at and making t the subject of the formula

$t=\frac {v-u}{a}$

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

$t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s$

Therefore, it needs 7.6 seconds to travel

7 0

3 years ago