Which type of seismic wave does not travel through liquid?

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

3 years ago

A 260-kg glider is being pulled by a 1,940-kg jet along a horizontal runway with an acceleration of a= 2.20 m/s^2 to the right.

lisov135 [29]

Answer:

a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

Explanation:

a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:

Jet

$\Sigma F = F - T = m_{J}\cdot a$ (1)

Glider

$\Sigma F = T = m_{G}\cdot a$ (2)

Where:

$F$ - Thrust of jet engines, measured in newtons.

$T$ - Tension in the cable connecting the jet and glider, measured in newtons.

$m_{G}$ , $m_{J}$ - Masses of the glider and the jet, measured in kilograms.

$a$ - Acceleration of the glider-jet system, measured in meters per square second.

If we know that $m_{G} = 260\,kg$ , $m_{J} = 1,940\,kg$ and $a = 2.20\,\frac{m}{s^{2}}$ , then the solution of this system of equations:

By (2):

$T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$T = 572\,N$

By (1):

$F = T+m_{J}\cdot a$

$F = 572\,N+(1,940\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$F = 4840\,N$

The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

7 0

3 years ago

During the spin cycle the time dependent angular speed of a washing machine drum is given by the equation ω(t) = at + bt2 - ct4

anastassius [24]

Answer:

α = 2.6 +1.7 t - 0.14 t³

Explanation:

Given that

ω(t) = at + bt² - ct⁴

where a = 2.6 rad/s², b = 0.85 rad/s³ and c = 0.035 rad/s⁵

We know that angular acceleration is the rate of change of angular velocity

α = dω/dt

ω(t) = at + bt² - ct⁴

dω/dt= a + 2 b t - 4 ct³

So

α = a + 2 b t - 4 ct³

By putting the values of a b and c

α = a + 2 b t - 4 ct³

α = 2.6 + 2 x 0.85 t - 4 x 0.035 t³

α = 2.6 +1.7 t - 0.14 t³

5 0

3 years ago

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Violet light of wavelength 405 nm ejects electrons with a maximum kinetic energy of 0.890 eV from a certain metal. What is the b

AleksandrR [38]

Answer:

Ф = 2.179 eV

Explanation:

This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.

E = K + Ф

the energy of the photons is given by the Planck relation

E = h f

we substitute

h f = K + Ф

Ф= hf - K

the speed of light is related to wavelength and frequency

c = λ f

f = c /λ

Φ = $\frac{hc}{\lambda } - K$

let's reduce the energy to the SI system

K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J

calculate

Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹

Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹

Ф = 3.4571 10⁻¹⁹ J

we reduce to eV

Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

Ф = 2.179 eV

4 0

3 years ago

Which would be the best way to represent the concentration of a 1.75 M K2CrO4 solution

AlekseyPX

1.75 moles per litre K2CrO4

7 0

4 years ago

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