The force needed to move an object 14 meters with 83 joules of work is what?

PLEASE HELP.. I WOULD BE SO HAPPY!! ILL MAKE YOLO THE BRAINLIEST....

madreJ [45]

Answer:

YOLOOOoo

Explanation:

6 0

3 years ago

An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq

wolverine [178]

Answer: a) 1 b) 61 W c) 61 W

Explanation:

a) The Power Factor (also known as cos φ), is defined by the difference in phase between current and voltage, in a RLC series circuit, and is expressed as follows:

cos φ = R / Z = R / $\sqrt{(R)^{2} + (Xl -Xc)^{2} }$

In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

This means that in resonance, current and voltage are in phase each other.

b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

7 0

3 years ago

During the storm a tree fell over into a river what might happen to the tree

mart [117]

The tree might get swept away by the current and it will disappear when it catches on something

4 0

3 years ago

A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu

kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) = ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i) work required to pump the contents to the rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the cone's rim

Extra work required = 28,914.229 lb × 5ft = 144571.145 lb.ft

Total work required = (281,913.733 + 144571.145) lb.ft

= 426,484.878 lb.ft

5 0

3 years ago

A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tal

olganol [36]

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance = 3.6 m

v = Image distance

f = Focal length = 35 mm

$h_u$ = Object height = 1.8 m

a) Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm$

The CCD sensor is 35.34 mm from the lens

b) Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}$

$m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm$

The person appears 17.67 mm tall on the sensor

7 0

3 years ago