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777dan777 [17]
3 years ago
12

SOMEONE HELP ME PLSS​

Physics
2 answers:
lara31 [8.8K]3 years ago
7 0
1. 2500/60 joules/sec
2. 2,500Nm
dsp733 years ago
4 0

Answer:

Explanation:

1 minute =60 second

power=work done/time taken

=2500/60

=41.66 watt

work=force *displacement

=500 N * 100 m

=50000 joule

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Difference between dry cell and simple cell​
lakkis [162]

Simple cells have liquid chemicals, making it harder for it to carry. While as dry cells have no liquid chemicals, making it easier to carry.

4 0
3 years ago
Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of
Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

V = 21\% * 0.66L

V = 0.1386L

Our values are given as

P = 1atm

T=293K R = 0.083145kJ*mol^{-1}K^{-1}

Using the equation to find n, we have:

PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

7 0
3 years ago
The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

distance ( d ) = 11.6 meter                                       ( m = mass of the car )

Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0
3 years ago
Science Net Forces. Could somebody help me?
tensa zangetsu [6.8K]
2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.

3) Balanced. Both boys are pulling with the same force. Neither is winning.

4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.

5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.

6) Balanced. You and the dog are pulling with the same force.
7 0
2 years ago
how much force is needed to move a brick with a mass of 345-kg a distance of 28m while doing 1008 j of work?
Travka [436]

    Work = (force) x (distance)

     1,008 J  =  (force) x (28 m)

Divide each side by 28m  :    (1,008 kg-m²/sec²) / (28 m)  =  force

                                           Force =  36 kg-m/s²  =  36 Newtons .
                                                                      (about 8.1 pounds)

It doesn't matter what that force accomplishes.
It could be moving a brick, lifting a fish, or pushing a little red wagon.
In order to do 1,008 joules of work in 28 meters, it takes 36 N of force,
in the direction of the 28 meters.
5 0
3 years ago
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