Combo Circuits VIR Charts. Can someone please fill this in?

Which substance—wood or steel—is the better conductor? A conductor is a material that requires very little

bogdanovich [222]

Steel is the correct answer because steel when on being applied heat will either change its state that too after hours because of its melting point being high or remain in that position only whereas wood turns into ashes...that is why we use metal utensils in the kitchen

7 0

4 years ago

a bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s² for 10s. Find the distance moved by it during this interval o

BaLLatris [955]

Answer:

90 m

Explanation:

Acceleration, $a=\frac {v-u}{t}$ where v and u are final and initial velocities respectively, t is the time taken

Substituting $1 m/s^{2}$ for a, 4 m/s for u and 10 s for t then

1*10=v-4

v=14 m/s

From kinematic equations

$v^{2}=u^{2}+2as$

Making s the subject then

$s=\frac {v^{2}-u^{2}}{2a}=\frac {14^{2}-4^{2}}{2\times 1}=90 m$

6 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0

3 years ago

How much force is required to accelerate a mass of 12kg at 3.0 m/s ^ 2

konstantin123 [22]

This is the answer your welcome

8 0

2 years ago

Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now i

prohojiy [21]

Answer:

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times8\times5.52\times10^{-4}}{1.99\times10^{-2}}$

$C=1.96\times10^{-12}$

We need to calculate the final capacitance

Using formula of capacitance

$C'=\dfrac{\epsilon_{0}A_{1}}{d}+\dfrac{k\epsilon_{0}A_{2}}{d}$

Put the value into the formula

$C'=(\dfrac{8.85\times10^{-12}}{1.99\times10^{-2}})((4.4\times5.52)+(3.6\times5.52)2.6)\times10^{-4}$

$C'=3.37\times10^{-12}$

We need to calculate the ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

$\dfrac{E}{E'}=\dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}C'V^2}$

Put the value into the formula

$\dfrac{E}{E'}=\dfrac{1.96\times10^{-12}}{3.37\times10^{-12}}$

$\dfrac{E}{E'}=0.58$

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

4 0

4 years ago