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Andrews [41]
3 years ago
5

What is meant by error​

Physics
1 answer:
Romashka [77]3 years ago
3 0

Answer:

An error (from the Latin error, meaning "wandering") is an action which is inaccurate or incorrect. In some usages, an error is synonymous with a mistake. In statistics, "error" refers to the difference between the value which has been computed and the correct value.

hope it will help you

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What is the force the when does when Gravity pushes you​
BARSIC [14]

Answer:

Image result for what is the force the when does when Gravity pushes you​

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

Explanation:

Image result for what is the force the when does when Gravity pushes you​

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

6 0
3 years ago
Read 2 more answers
Quanto vale a resultante de duas forças de mesmo módulo mesma direção e sentidos opostos atuando sobre um corpo de massa igual a
Crank
A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.
6 0
3 years ago
Read 2 more answers
a plane passes over Point A with a velocity of 8,000 m/s north. Forty seconds later it passes over Point B with a velocity of 10
ValentinkaMS [17]

Acceleration = (change in velocity) / (time for the change)

Change in velocity = (ending velocity) - (starting velocity)

Change in the plane's velocity = (10,000 m/s north) - (8,000 m/s north)

Change in the plane's velocity = 2,000 m/s north

Time for the change = 40 seconds

Acceleration = (2,000 m/s north) / (40 seconds)

<em>Acceleration = 50 m/s² north </em>

8 0
3 years ago
Enunciado del ejercicio n° 1
hoa [83]

Answer:

34

Explanation:

6 0
3 years ago
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
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