Ask question

Ask question

All categories

2 years ago

8

6. Given a sheet of metal that is 1.2 cm wide, 3.8 cm long and 1.0 mm thick with a density of 8.57 g/cm3, calculate the mass of

the metal.

1 answer:

Jobisdone [24]2 years ago

5 0

Answer:

39.08 or 39.0792

Explanation:

1.2*3.8=4.56 8.57*4.56=39.0792 (Round up is 39.08

Volume: m/v*d (The m over v goes as a fraction and the density is in the middle beside the m over v with a multiplication sign)

You might be interested in

What causes a boat to float? My questions are super easy. So you can follow me if you like easy question.

marin [14]

Boats float because the gravity is acting down on it and the buoyant force is acting up on the ship.

6 0

3 years ago

Read 2 more answers

A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?

Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

8 0

3 years ago

Use the graph to determine which statement is true about the end behavior of f(x).

Airida [17]

Answer:

As the x-values go to negative infinity, the function’s values go to positive infinity.

Explanation:

if the ans choices are:

As the x-values go to negative infinity, the function’s values go to negative infinity.

As the x-values go to negative infinity, the function’s values go to positive infinity.

As the x-values go to positive infinity, the function’s values go to negative infinity.

As the x-values go to positive infinity, the function’s values go to zero.

the ans is the 2nd choice

4 0

2 years ago

Read 2 more answers

Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5

Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0

2 years ago

A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.

Lena [83]

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, $\mu = 2.501 V$

standard deviation, $\sigma = 0.113 V$

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, $P(X\leq 2.70)$ :

$z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761$

Now, from the Probability table of standard normal distribution:

$P(z\leq 1.761) = 0.9608$

Now,

$P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%$

Now, for the expected no. of readings greater than 2.70 V:

$P(X\geq 2.70) = \frac{No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V}{Total\ no.\ of\ readings}$

No. of readings expected to be greater than 2.70 V = $P(X\geq 2.70)\times Total\ no.\ of\ readings$

No. of readings expected to be greater than 2.70 V = $0.0392\times 60 = 2.352$ ≈ 2

7 0

3 years ago