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3 years ago

PLEASE HELP ME

Physics

1 answer:

tigry1 [53]3 years ago

5 0

Our lives would big be possible because we would not be able to drive cars all knots would untie

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How old is a bone if it still has 50% of its carbon-14 content?

Degger [83]

Answer:

5730 years

Explanation:

The half life of carbon-14 is 5730 years. If 50% of the carbon-14 remains, then exactly 1 half life has passed.

The half-life equation is:

A = A₀ (½)^(t / T)

where A is the remaining amount,

A₀ is the initial amount,

t is time,

and T is the half life.

In this case, A = ½ A₀ and T = 5730.

½ A₀ = A₀ (½)^(t / 5730)

½ = (½)^(t / 5730)

1 = t / 5730

t = 5730

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2 years ago

The potential difference between the plates of an ideal air-filled parallel-plate capacitor with a plate separation of 6.0 cm is

Yanka [14]

Answer:

1000 N/C

Explanation:

Potential difference, V = 60 V

Distance between the plates, d = 6 cm = 0.06 m

The electric field between the plates is given by

E = V / d

E = 60 / 0.06 = 1000 N/C

Thus, the electric filed between the plates is 1000 N/C.

8 0

3 years ago

In a digital multimeter, what changes inside the meter when you push the button to change from 200v range to the 20v range? by w

charle [14.2K]

So we want to know what changes inside the multimeter when we change the voltage range from 200 V to 20 V, by what factor and does it increase or decrease. What we want when trying to measure the voltage with a multimeter is that a minimal current passes trough the mulitmeter so when we change the voltage range, we decrease the resistance by a factor of 10 because the voltage is decreased by a factor of 10.

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3 years ago

A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface

Fiesta28 [93]

Answer:

$X_2=25.27m$

Explanation:

Here we will call:

1. $E_1$ : The energy when the first spring is compress

2. $E_2$ : The energy after the mass is liberated by the spring

3. $E_3$ : The energy before the second string catch the mass

4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. $E_1 = E_2$

2. $E_2 -E_3= W_f$

3. $E_3 = E_4$

where $W_f$ is the work of the friction.

1. equation 1 is equal to:

$\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2$

where K is the constant of the spring, x is the distance compressed, M is the mass and $V_2$ the velocity, so:

$\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2$

Solving for velocity, we get:

$V_2$ = 65.319 m/s

2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so: