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2 years ago

A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.

Physics

1 answer:

jeka942 years ago

6 0

Mass of the elevator (m) = 570 Kg
Acceleration = 1.5 m/s^2
Distance (s) = 13 m
Let the force be F.
We know, F = ma,
Therefore, F = (570 × 1.5) N = 855 N
Angle between distance and force (θ) = 0°
We know, work done = F s Cos θ
Therefore, work done by the cable during this part
= (855 × 13 × Cos 0°) J
= (855 × 13 × 1) J
= 11115 J

<u>Answer</u><u>:</u>

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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

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3 years ago

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zlopas [31]

its C because i just anwserd it & it was right

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3 years ago

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zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

$x=v_0\cos20.0^\circ t+\dfrac12a_xt^2$

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The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

$x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ$

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