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2 years ago

What noncontact forces act on an object that are balanced

Physics

1 answer:

Firlakuza [10]2 years ago

3 0

Answer:

I can give you a list of noncontact forces off of my head, if that is what you need.

Explanation:

First, gravity. It acts everywhere and pulls down every object that is in the air towards the ground, unless they have an equal and opposite force that counters it.

Second, wind. I don't think this will count because wind is a physical thing, but it can be used in the same way as we did with gravity. If the force counteracting it is larger, equal, and opposite of the force it is countering, then it will move forward, through the wind.

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What is the acceleration of a 10kg pushed by a 5n force

Semenov [28]

Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a

8 0

3 years ago

A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person

vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 $\frac{m}{s}$

t=0.475s V=1.95 $\frac{m}{s}$

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

$sin(\alpha) =\frac{op}{h}$

$op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m$

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

$T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz$

c).

$T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz$

The period is the inverse of the time of the motion so, the T1 is faster that the T because

$t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s$

d).

The speed is the relation between the distance with time so:

$Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}$

3 0

2 years ago

17. Matthew has a piece of wood with mass of 278 g and a volume of 375 cm^3. What is the wood’s density? a. Step 1: Formula ____

Keith_Richards [23]

Answer:

Density =0.74 $g/cm^{3}$

Explanation:

Step 1: Formula

$Density=\frac{Mass}{Volume}$

Step 2: Data

m=278 g

v=375 $cm^{3}$

Step 3:Solve

d= $\frac{278}{375}$ $g/cm^{3}$

d=0.74 $g/cm^{3}$

8 0

2 years ago

The water cycle was described in the Scriptures long before modern scientists understood it.

s344n2d4d5 [400]

Answer:

False

Explanation:

Low humidity is typically associated with cold temperatures since cold air holds less water compared to warm air. While it is possible for indoor humidity to be as low as 10%, most people generally find humidity levels of about 40 to 45 percent comfortable.

4 0

2 years ago

Read 2 more answers

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240

Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

$V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC$

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0

2 years ago