What is the mass of an object that requires a force of 30 N to accelerate at a rate of 5 m/sec2 ?

(a) Calculate the self-inductance (in mH) of a 55.0 cm long, 10.0 cm diameter solenoid having 1000 loops.

DedPeter [7]

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

$L=\dfrac{\mu_o N^2A}{l}$

A is area

$L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH$

(b) The energy stored in the inductor is given by :

$E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J$

Hence, this is the required solution.

4 0

3 years ago

suppose a ball is thrown vertically upward. Eight seconds later it returns to its point of release. What is the initial velocity

valentinak56 [21]

The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.

At its highest point, its velocity changed from upward to downward.
At that instant, its velocity was zero.

The acceleration of gravity is 9.8 m/s². That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second.

-- If the object is falling downward, it moves 9.8 m/s faster every second.

-- If the object is tossed upward, it moves 9.8 m/s slower every second.

The ball took 4 seconds to lose all of its upward speed. So it must have
been thrown upward at (4 x 9.8 m/s) = 39.2 m/s .

(That's about 87.7 mph straight up. Somebody had an amazing pitching arm.)

6 0

3 years ago

sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he

podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

5 0

3 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )