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3 years ago

Which answer choice provides the best set of labels for Wave A and Wave B?

Physics

2 answers:

Blizzard [7]3 years ago

4 0

Answer:

Answer Choice D:

Wave A: Low Frequency Wave

Wave B: High Frequency Wave

Explanation:

Lower Frequency waves have a higher wavelength, while higher frequency waves have a lower wavelength.

hodyreva [135]3 years ago

3 0

A has less energy and lower frequency, while B has greater energy and higher frequency.

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Which group of items all contain a transparent part with at least one curved surface to refract light?

salantis [7]

The answer is going to be a

5 0

3 years ago

A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface a

kykrilka [37]

Answer:

v₀ = 280.6 m / s

Explanation:

we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,

We write the mechanical energy when the shock has passed the bodies

Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

$Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\ Em_0 = Em_{f}$

½ (m + M) v² = ½ k x²

Let's calculate the system speed

v = √ [k x² / (m + M)]

v = √[152 ×0.78² / (0.012 +0.109) ]

v = 27.65 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

p₀ = m v₀

After the crash

$p_{f} = (m + M) v$

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

$p_0 = p_{f}$

m v₀ = (m + M) v

v₀ = v (m + M) / m

let's calculate

v₀ = 27.83 (0.012 +0.109) /0.012

v₀ = 280.6 m / s

4 0

3 years ago

A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the

Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g= $\frac{530}{1000}=0.530 kg$

1kg=1000 g

Area=A= $935 cm^2=935\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Height,h=4.4 m

C=1

$\rho=1.21 kg/m^3$

Width of rectangular prism,b=11.6 cm= $\frac{11.6}{100}=0.116 m$

1 m=100 cm

Length,l=23.2 cm= $0.232 m$

Area= $l\times b=0.116\times 0.232=0.0269 m^2$

Terminal velocity, $v_t=\sqrt{\frac{2mg}{\rho CA}}$

Where $g=9.8 m/s^2$

Using the formula

$v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}$

$v_t=17.86 m/s$

The velocity of person, $v=\sqrt{2gh}$

Using the formula

$v=\sqrt{2\times 9.8\times 4.4}$

$v=9.29 m/s$

4 0

2 years ago

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b. $v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

c.Work don=Final K.E-Initial K.E

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

4 0

2 years ago

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Inside a television picture tube there is a build-up of electrons (charge of 1.602 × 10^–19 C) with an average spacing of 38.0 n

Brut [27]

The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9 N. m2 / C2

Given are the following:
Q = 1.602 × 10^–19 C
r = 38 x 10^-9 m

Substitue the given:
E = $\frac{( 1.602 x 10^{-19} )( 9.0x10_{9} )}{(38x10^{-9}) ^{2} }$

E = 998.476 kN/C

8 0

3 years ago

Read 2 more answers