3 years ago

Describe a situation where you could properly conclude that a set of data is precise but

Physics

1 answer:

Orlov [11]3 years ago

4 0

It’s not arrested because if you look closer it’s plug walk I don’t even understand how it talk tho

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Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x.

Contact [7]

The center of mass is given with this formula:
$x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}$
Velocity is:
$v=\frac{dv}{dt}$
So, for the velocity of the center of mass we have:
$\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\$
In our case it is:
$v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}$

5 0

3 years ago

Answers to electrical?

Orlov [11]

Can you get a better pic so i can read it

3 0

3 years ago

If you have 10.0 g of a substance that decays with a half-life of 14 days, then how much will you have after 42 days?

Nat2105 [25]

The formula for the mass that remains:
$m=m_0 \times (\frac{1}{2})^\frac{t}{T}$
m₀ - the initial mass, t - time, T - the half-life

$m_0=10 \ g \\
T=14 \ d \\
t=42 \ d \\ \\
m=10 \times (\frac{1}{2})^\frac{42}{14}=10 \times (\frac{1}{2})^3=10 \times \frac{1}{8}=10 \times 0.125=1.25$

The answer is c. 1.25 g.

6 0

3 years ago

Would u rather/ be able to fly. or be able to turn invisable

mojhsa [17]

Answer:

fly......................

4 0

2 years ago

Read 2 more answers

A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the

Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)

v₀ is the initial voltage on the cap

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

τ = 20µ x 1.5k = 30 ms

v = v₀e^(t/τ)

0.5 = 1.5e^(t/30ms)

e^(t/30ms) = 10/3

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0

3 years ago