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Salsk061 [2.6K]
3 years ago
9

How can you tell when the forces acting on a system are balanced?

Physics
1 answer:
mojhsa [17]3 years ago
5 0

Answer:

a. no lift

Explanation:

To determine if the forces acting upon an object are balanced or unbalanced, an analysis must first be conducted to determine what forces are acting upon the object and in what direction. If two individual forces are of equal magnitude and opposite direction, then the forces are said to be balanced.

plz give me a brianliest

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If a 10. m3 volume of air (acting as an ideal gas) is at a pressure of 760 mm and a temperature of 27 degrees Celsius is taken t
kow [346]
We know, the ideal gas equation, 
P1V1 / T1 = P2V2 / T2

Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K

P2 = 400 mm Hg
T2 = -23 + 273 = 250 K

Substitute their values, 
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83

In short, Your Answer would be approx. 15.83 m3

Hope this helps!
7 0
3 years ago
A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea
Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
R_T = R_1 + R_2 + ... + R_n

In parallel:
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}

2.

We can use Ohm's law to solve for the current in the circuit.

i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

V = iR\\\\V = (6)(15) = \boxed{90 V}

4 0
2 years ago
15. You are watching a baseball game on television that 15
Gre4nikov [31]

Answer:

t_total = 6.99 s

Explanation:

It asks us how long it takes to hear the sound, for this we must look for the time (t₁) it takes for the sound to reach the microphone, the time it takes for the video signal (t₂) to reach the television and the time (₃) it takes for the TV sound to reach us, so the total delay time is

         t_total = t₁ + t₂ + t₂

we look for t1, it indicates that the distance x = 22m

          v = x / t

           t = x / v

           t₁ = 22/343

           t₁ = 6.41 10-2 s

time t₂

          t₂ = 4500 103/3 108

          t₂ = 1.5 10-5 s

time t₃

          t₃ = 2/343

          t₃ = 5.83 10⁻³

Total time is

         t_total = t₁ + t₂ + t₃

         t_total = 6.41 10⁻² + ​​1.5 10⁻⁵ + 0.583 10⁻²

         t_total = 6.99 s

5 0
3 years ago
A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th
forsale [732]

Answer:

F=-18412.9N, where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

1\ hour = 3600\ seconds

1\ mile = 1600\ meters

1000g = 1kg

We can write that:

\frac{1\ hour}{3600\ seconds}=1

\frac{1600\ meters}{1\ mile}=1

\frac{1kg}{1000g}=1

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s

110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s

145 g=145 g(\frac{1kg}{1000g})=0.145kg

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is a=\frac{\Delta v}{\Delta t}, so we have:

F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}

Taking the throw direction as the positive one, for our values we have:

F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N

4 0
3 years ago
The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space
Svet_ta [14]
D not 100% sure though.
4 0
3 years ago
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