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2 years ago

An air mass that forms over Arizona and New Mexico will be?

Physics

1 answer:

vlada-n [284]2 years ago

5 0

Continental Tropical Air Mass. Your welcome matey

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A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

4 0

3 years ago

Which type of nuclear waste can be disposed of in a landfill?

marysya [2.9K]

Low-level nuclear waste can be disposed of in a landfill.

8 0

2 years ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

$E_{k1}+W=E_{k2}\\$

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

$\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]$

We know that work is defined as the product of force by distance.

$W=F*d$

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

$f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]$

Now solving the equation for the work.

$d=W/F\\d = 50/19.62\\d = 2.54[m]$

4 0

2 years ago

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib

Natasha_Volkova [10]

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

$L_1A_1 = L_2A_2$