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ANTONII [103]
3 years ago
12

Mindy places a strip of zinc in a copper sulfate solution, as shown in the

Physics
2 answers:
mr Goodwill [35]3 years ago
8 0

Answer:

C. The concentration of the copper sulfate is too low

Explanation:

slamgirl [31]3 years ago
6 0

Answer:

The concentration of copper sulfate is too low

Explanation:

Ap3x

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A 55 kg person falling with a velocity of 0.6
siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

3 0
2 years ago
A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has
lara [203]

Answer:

The phase constant is 7.25 degree  

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = \sqrt{(\frac{v_x}{\omega })^2+y^2 }  = \sqrt{(\frac{35}{2\times \pi  \times y})^2+6.2^2 }  =  6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ  = 0.992

Ф = 7.25 degree  

so the phase constant is 7.25 degree  

5 0
3 years ago
When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the
Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

Voltage drop across the wire V = 0.55 volt

We know that resistance is given by R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega

Length of the copper wire l = 1 m

Resisitivity of the copper wire \rho =1.72\times 10^{-8}\Omega m

We know that resistance R=\frac{\rho l}{A}

0.00177=\frac{1.72\times 10^{-8}\times  1}{A}

A=969.4545\times 10^{-8}m^2

As area A=\pi r^2

3.14\times r^2=969.4545\times 10^{-8}

r=17.57\times 10^{-4}m

So diameter d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m = 351.42 mm

3 0
3 years ago
A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle
Yuri [45]

Answer:

(a). Index of refraction are n_{red} = 1.344 & n_{violet} = 1.406

(b). The velocity of red light in the glass v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

The velocity of violet light in the glass v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

Explanation:

We know that

Law of reflection is

n_1 \sin\theta_{1} = n_2 \sin\theta_{2}

Here

\theta_1 = angle of incidence

\theta_2 = angle of refraction

(a). For red light

1 × \sin 56.6 = n_{red} × \sin 38.4

n_{red} = 1.344

For violet light

1 × \sin 56.6 = n_{violet} × \sin 36.4

n_{violet} = 1.406

(b). Index of refraction is given by

n = \frac{c}{v}

n_{red} = 1.344

v_{red} = \frac{c}{n_{red} }

v_{red} = \frac{3(10^{8} )}{1.344}

v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

v_{violet} = \frac{3(10^{8} )}{1.406}

v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

This is the velocity of violet light in the glass.

8 0
3 years ago
A glass lens that has an index of refraction equal to 1.57 is coated with a thin layer of transparent material that has an index
Bogdan [553]

Answer:

Find the given attachment

5 0
3 years ago
Read 2 more answers
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