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3 years ago

What noncontact forces act on an object that are balanced

Physics

1 answer:

Firlakuza [10]3 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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The difference between an experimental value and an accepted value is

likoan [24]

Answer:

HERE'S MY UNDERSTANDING OF THE DIFFERENCE

3 0

3 years ago

A boy of mass 80 kg slides down a vertical pole, and a frictional force of 480 N acts on him. What is his acceleration as he sli

olga_2 [115]

Answer:

His acceleration is $\overrightarrow{a}=4\frac{m}{s^{2}}$

Explanation:

Newton's second law states that acceleration of a body is cause by a net force, the relation between them is:

$\sum\overrightarrow{F}=m\overrightarrow{a}$

On the boy there're acting two forces, his weight (W) that points downward and the frictional force (f) that points upward (they boy moves downward and friction always is opposite to movement). So $\sum\overrightarrow{F}=\overrightarrow{W}+\overrightarrow{f}$ so (1) is:

$\overrightarrow{W}+\overrightarrow{f}=m\overrightarrow{a}$

Using the positive direction downward weight and gravitational acceleration(g) are positive and friction force is negative:

$W-f=m\overrightarrow{a}$ , solving for a:

$\overrightarrow{a}=\frac{W-f}{m}$ , weight is mg:

$\overrightarrow{a}=\frac{mg-f}{m}=\overrightarrow{f}=\frac{(80kg)(10\frac{m}{s^{2}})-(480)}{80}$

$\overrightarrow{a}=4\frac{m}{s^{2}}$

6 0

4 years ago

If an atom gains or loses a valence electron, the atom becomes a charged particle called a/an_____________.

erica [24]

D i think probably right

5 0

3 years ago

Read 2 more answers

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

6 0

3 years ago

On a spending spree in Malaysia, you buy an ox with a weight of 28.9 piculs in the local unit of weights: 1 picul = 100 gins, 1

Talja [164]

Answer:

1747.41 kg

Explanation:

If 1 piculs = 100 gins,

Then 28.9 piculs = (28.9 × 100) gins = 2890 gins.

Also,

If 1 gin = 16 tahils

then 2890 gins =( 16 × 2890 ) = 46240 tahils

Also,

If 1 tahil = 10 chees,

Then 46240 tahils = 46240 × 10 chees = 462400 chees.

Also,

If 1 chee = 10 hoons

Then 462400 chees =( 10 × 462400) hoons = 4624000 hoons.

Also,

If 1 hoon = 0.3779 g,

Then 4624000 hoons = 0.3779 × 4624000 = 1747409.6 g.

The mass in kg that i will declare in the shipping manifest = (1747409.6/100 )kg = 1747.41 kg

6 0

3 years ago