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2 years ago

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Speed velocity and acceleration escape room

1 answer:

Svetllana [295]2 years ago

7 0

Answer:

The speed velocity and acceleration escape room is an immersive experience for your students. It allows them to demonstrate their knowledge of these principles in a fun and engaging way. Speed Velocity Acceleration Escape Room - Single Student DIGITAL Version

Explanation:

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1) La longitud del brazo de potencia de una palanca es de 0,0035 hectómetros y la del brazo de resistencia es de 55 centímetros.

Arlecino [84]

Answer:

Entonces seria 127 para vencer.

Explanation:

espero averte ayudado:-)

6 0

3 years ago

Tony waits for the spring tide to bring the best surfing waves in his area. He knows that the moon affects the tides, so which m

Komok [63]

Answer:

c

Explanation:

just looked it up

8 0

3 years ago

Read 2 more answers

Your laundry basket weighs 22 N and your room is 3.0 m above you on the second floor. It takes you 6.0 seconds to carry the laun

Fofino [41]

<h2>Power is 11 W</h2>

Explanation:

Power = Work ÷ Time

Work = Force x Displacement

Force = 22 N

Displacement = 3 m

Time = 6 seconds

Substituting

Work = Force x Displacement

Work = 22 x 3 = 66 J

Power = Work ÷ Time

Power = 66 ÷ 6

Power = 11 W

Power is 11 W

8 0

3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

2 years ago

How long does it take for an airplane to change its velocity from 140 m/s to 180 m/s if its

kobusy [5.1K]

Answer:

Explanation:

a = 4ms⁻², Vf = 180 m/s & Vi = 140m/s

a = $\frac{Vf-Vi}{t}$

4 = $\frac{180-140}{t}$

t = 40/4

t = 10sec

To Measure Distance Use third Equation of Motion:

2aS = Vf²-Vi²

S = $\frac{180*180 - 140*140}{2(4)}$

S = 12800/8 = 1600m

4 0

3 years ago