Physics Middle School

Which of he following injuries include instability of the knee with pain and swelling?

Answers

Answer 1

Hi there,

the following injuries that include instability of the knee with pain and swelling is ACL Tear.

The ACL tear is common especially in sports like football one wrong move and you fall and tear your ACL

Hope this helped :)

Have a great day

Related Questions

In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to accelerate from 18.0m/s to 30.0m/s. What is the magnitude of the car's acceleration?

Answers

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²

Final answer:

The magnitude of the car's acceleration is calculated using the change in velocity over the time taken. With a change in velocity of 12.0 m/s over 8.0 seconds, the car's acceleration is 1.5 m/s².

Explanation:

To find the magnitude of the car's acceleration, we can use the formula for acceleration, which is the change in velocity (Δv) divided by the time (Δt) it takes for that change. In this case, the car's velocity increases from 18.0 m/s to 30.0 m/s over 8.0 seconds.

We calculate acceleration (a) as follows:

a = Δv / Δt

Δv = final velocity - initial velocity = 30.0 m/s - 18.0 m/s = 12.0 m/s

Δt = 8.0 s

Thus, a = 12.0 m/s / 8.0 s = 1.5 m/s².

The magnitude of the car's acceleration is 1.5 m/s².

Calculate the momentum for a 0:2 kg rifle bullet traveling 300m/a

Answers

Given data

          mass (m) = 0.2 Kg ,

       velocity (v) =300 m/s ,

      calculate momentum (p) = m × v  

                                                = 0.2 × 300

                                              P  = 60 Kg. m/s

Momentum of the bullet is 60 Kg. m/s..

The power in a lightbulb is given by the equation P- FR where /is the current flowing through the lightbulb and Ris the resistance of the lightbulb. What is the current in a circuit that has a resistance of 30.0 o and a power of 2.00 W?

A.) 15.0 A
B.) 3.87 A
C.) 0.258 A
D.) 0.067 A

(PLEASE HELP NEED ANSWER ASAP)

Answers

POWER OF CIRCUIT

P=VI

While,

VOLTAGE OF CIRCUIT

V=IR

THEN,

P=I^R

P/R=I^

2.00/30.0=I^

I=0.258A.

Answer:0.258

Explanation:

an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respectively. if the energy input is 3600 joule. what is the work done by the engine

Answers

Given data

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

                We know that, efficiency (η) = Net work done ÷ Heat supplied

                                                           η =   W ÷ Q₁  

                                                           W = η × Q₁

               First determine the efficiency ( η ) = ?

                                Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

                                                                        = 33.3% = 0.333

               Now, Work done is W = η × Q₁

                                                    = 0.33 × 3600

                                                   W = 1188 J

Work done by the engine is 1188 J

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