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3 years ago

Burl and Paul have a total weight of 992 N. The tensions in the ropes that

Physics

1 answer:

exis [7]3 years ago

6 0

Answer:

1670 N

Explanation:

The weight of ropes is negligible hence considering that Burl and Paul have a total weight of 992 N while the total weight of the scaffold while they are in it is equivalent to 2662 N then to get the weight if scaffold we have to deduct the the weight of Burl and Paul from the total weight.

Therefore, scaffold will be 2662-992= 1670 N

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Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.

tresset_1 [31]

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, $L=50\ mH=0.05\ H$

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

$X_L=2\pi fL$

$X_L=2\pi \times 400\ Hz\times 0.05\ H$

$X_L=125.66\ \Omega$

$X_L=125.7\ \Omega$

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

8 0

3 years ago

White blood cells are most closely associated with which two body’s systems

Vlad1618 [11]

Answer:

Explanation:

circulatory and digestive. ... An immune response is primarily due to the body's white blood cells recognizing.

8 0

2 years ago

Help me, Ive been stuck for 15 minutes

natita [175]

The answer is b) venus because the period rotation is also referred to as a day so if venus’ period of rotation is 243 [earth] days and its year is 225 [earth] days, then the period of rotation is longer.

5 0

2 years ago

A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.

charle [14.2K]

Answer: $0.00024\ m/s^2$

Explanation:

Given

Radius of flywheel is $r=0.4\ mm$

Angular acceleration $\alpha=0.6\ rad/s^2$

For no change in radius, tangential acceleration is given as

$\Rightarrow a_t=a\lpha \times r$

Insert the values

$\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2$

5 0

2 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

7 0

3 years ago