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Kobotan [32]
3 years ago
14

What simple machines would move or turn an object?

Physics
1 answer:
Vsevolod [243]3 years ago
5 0
Its a lever If I'm not mistaking
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A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration
kiruha [24]
Average acceleration  =  (change in speed) / (time for the change) .
  
Change in speed = (ending speed) - (beginning speed)

                       =  (9.89 miles/hour) - (2.35 yards/second)  = 26,839.2 ft/hr

Acceleration  =  (26,839.2 ft/hr) / (4.67 days)  =  2,873.58 inch/hour²  
6 0
3 years ago
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
3 years ago
Saatlemati
vodomira [7]

Answer:

what are u asking there isnt a question

3 0
3 years ago
The leg's force forward on the foot= 500N
Anna11 [10]

There's so much going on here, in a short period of time.

<u>Before the kick</u>, as the foot swings toward the ball . . .

-- The net force on the ball is zero.  That's why it just lays there and
does not accelerate in any direction.

-- The net force on the foot is 500N, originating in the leg, causing it to
accelerate toward the ball.


<u>During the kick</u> ... the 0.1 second or so that the foot is in contact with the ball ...

-- The net force on the ball is 500N.  That's what makes it accelerate from
just laying there to taking off on a high arc.

-- The net force on the foot is zero ... 500N from the leg, pointing forward,
and 500N as the reaction force from the ball, pointing backward. 

That's how the leg's speed remains constant ... creating a dent in the ball
until the ball accelerates to match the speed of the foot, and then drawing
out of the dent, as the ball accelerates to exceed the speed of the foot and
draw away from it.


5 0
3 years ago
_ NaBr + __ H3PO4 → ___ Na3PO4 + __ HBr
algol [13]

Answer:

The balanced equation is 3NaBr+1H3PO4 ----> 1Na3PO4 + 3HBr

This is a double replacement because you are switching both the Na and the Hydrogen.

Explanation:

3 0
3 years ago
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