What's the opposite of futility

Sound waves cannot carry energy through. A water B air C a mirror D a vacuum

Ber [7]

I looked up the question and got D- a vacuum

3 0

3 years ago

A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an

Dominik [7]

Answer:

position as a function of time is y = 0.05 × cos(9.9)t

Explanation:

given data

mass = 5 kg

length = 10 cm = 0.1 m

displaced = 5 cm

to find out

position as a function of time

solution

we will apply here equilibrium that is

mass × g = k × length

put here value and find k

k = $\frac{5*9.8}{.01}$

k = 490 N/m

and ω is

ω = $\sqrt{\frac{k}{m} }$

ω = $\sqrt{\frac{490}{5} }$

ω = 9.9

so here position w.r.t time is

y = 0.05 × cosωt

y = 0.05 × cos(9.9)t

so position as a function of time is y = 0.05 × cos(9.9)t

8 0

3 years ago

How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will

nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg$

So, the mass that is attached to the spring is 1 kg.

6 0

3 years ago

The diagram below shows a wave with its wavelength indicated in red.

lisabon 2012 [21]

C. It will decrease

5 0

4 years ago

Read 2 more answers

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

3 years ago

What's the opposite of futility​

What's the opposite of futility