All categories

2 years ago

In an RL parallel circuit, VT = 240 V, R = 330 Ω, and XL = 420 Ω. What is the Apparent Power (VA)?

Engineering

1 answer:

Amiraneli [1.4K]2 years ago

6 0

Answer:

that answer is correct

Explanation:

This answer is correct because they explained everything they needed.

You might be interested in

Select the statement that is false.

ra1l [238]

Answer:

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.

8 0

2 years ago

How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?

nataly862011 [7]

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

= 20.4kg · 9.81 $\frac{N}{kg}$ · 1.50m

= 35.316 $\frac{N}{m}$

= 35.316 W

4 0

2 years ago

A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same inst

expeople1 [14]

Answer:

s= 20.4 m

Explanation:

First lets write down equations for each ball:

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s

now we just have to put that time in any of those equations an get distance from the ground:

s = 30 + 5*2 -1/2*9.81 *2^2

s= 20.4 m

6 0

2 years ago

Pie charts should have no more than eight segments. True or False?

Pepsi [2]

Answer:

Explanation:

Pie charts generally should have no more than eight segments.

6 0

2 years ago

Read 2 more answers

A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi

raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation $=\frac{N_1 +N_2}{2}$

$Nm = \frac{500+750}{2} = 625 rpm$

$w =\frac{2\pi Nm}{60}$

=65.44 rad/sec

$F_T = mw^2 e$

$F_T = mew^2$

= 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio $=\frac{300}{428.36} = 0.7$

also

transmission ratio $= \frac{1}{[\frac{w}{w_n}]^{2} -1}$

$0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}$

SOLVING FOR Wn

Wn = 42 rad/sec

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

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

3 0

3 years ago