2 years ago

Please Help with thia question

Physics

1 answer:

telo118 [61]2 years ago

6 0

Answer:

the types of force shown in the picture is

balanced force.

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Fenómeno químico en virtud del cual se transforma un cuerpo o compuesto por la accion de un oxidante

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Chemical phenomenon by virtue of which a body or compound is transformed by the action of an oxidant

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3 years ago

your friend sit in a sked in the snow. if you apply a force of 120 N to them, they have an acceleration of 1.3 m/s2. what is the

levacccp [35]

Need to know the equation for force
F=MA
F is force
M is mass- we need to know the mass
A is acceleration
use "x" for mass
120 N= 1.3x
divide 1.3 in both side
kg unit for mass
X=92.31 kg
or
mass = 92.31 kg
Hope this helps

7 0

3 years ago

A circuit is a continuous loop. a. True b. False

TiliK225 [7]

False its an unbroken loop that allows electrons to flow through constantly

4 0

3 years ago

Read 2 more answers

The sound from a loud speaker has an intensity level of 80 db at a distance of 2.0m. Consider the speaker to be a point source,

Tamiku [17]

Answer:

2.83m

Explanation:

The information that we have is

Intensity at 2.0 m: $I=80dB$ and $r_{1}=2m$

we need an intensity level of: $I_{2}=40dB$

thus, we are looking for the distance $r_{2}$ .

which we can find with the law for intensity and distance:

$(\frac{r_{2}}{r_{1}} )^2=\frac{I_{1}}{I_{2}}$

we solve for $r_{2}$ :

$\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }$

and we substitute the known values:

$r_{2}=(2m)\sqrt{\frac{80dB}{40dB} }\\\\r_{2}=(2m)\sqrt{2}\\ r_{2}=2.83m$

at a distance of 2.83m the intensity level is 40dB

5 0

3 years ago

A 67-kg base runner begins his slide into second base when he is moving at a speed of 3.6 m/s. The coefficient of friction betwe

USPshnik [31]

Answer:

434.16 Joules

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of person

From work-energy theorem

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}\times 67\times (3.6^2-0)\\\Rightarrow KE=434.16\ Joules$

The runner loses 434.16 Joules of mechanical energy.

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow \mu mg=\frac{W}{s}\\\Rightarrow s=\frac{W}{\mu mg}\\\Rightarrow s=\frac{434.16}{0.7\times 67\times 9.81}\\\Rightarrow s=0.94364\ m$

He slides 0.94364 m

4 0

3 years ago

Please Help with thia question​

Please Help with thia question