The periodic table arranges elements by increasing ________

An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. If the scaffold weighs 500 N and t

Arturiano [62]

Answer:

T = 850 N

Explanation:

given,

mass of billboard worker = 800 N

length of scaffold = 4 m

weight of the scaffold = 500 N

worker is standing at 1 m from one end.

Tension in the rope = ?

To calculate the tension in the string we have to balance the clockwise and counterclockwise moment of the system.

Weight of the worker and the weight of the scaffold will be in clockwise direction where as the tension will be in counterclockwise direction

now,

800 x 3 + 500 x 2 = T x 4

4T = 3400

T = 850 N

hence, tension in the rope is equal to 850 N

5 0

4 years ago

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential diffe

bazaltina [42]

To solve this problem we will apply the concepts related to the calculation of power, from the two electrical forms:

$P = VI$

$P = \frac{V^2}{R}$

Here,

V = Voltage

I = Current

R= Resistance

P = Power

PART A)

$R = \frac{V^2}{P}$

Replacing,

$R = \frac{(120V)^2}{150W}$

$R = 96\Omega$

Resistance of bulb is $96\Omega$

PART B)

$I = \frac{P}{V}$

$I = \frac{150W}{120V}$

$I = 1.25A$

The bulb will draw 1.25A current

7 0

3 years ago

When populations of two species work together to obtain resources, they are

valina [46]

<span>cooperation. i think that is it

</span>

7 0

3 years ago

Read 2 more answers

Can some one please help!!!!!!<br> ASAP pleaseeeeeeee❗️❗️❗️❗️❗️❗️❗️❗️

Thepotemich [5.8K]

a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

a)

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

The total energy when the spring is compressed is:

$E=KE_i + PE_{si}$

with

$KE_i = 0$ (initial kinetic energy is zero)

$PE_{si} = \frac{1}{2}kx^2$ is the elastic potential energy stored in the spring, with

k = 450 N/m (spring constant)

x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

$E=KE_f + PE_{sf}$

with

$KE_f = \frac{1}{2}mv^2$ (final kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = velocity of launch of the rocket

$PE_{sf} = 0$ (elastic potential energy is zero when the spring is released)

Combining the two equations we get

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

And solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(450)(0.10)^2}{0.15}}=5.48 m/s$

b)

In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

$E=KE_i + PE_{gi}$