What is the primary source of a sound?

Look at the diagram. What is the net force acting on the object?

nlexa [21]

The answer would be A

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%5Csqrt%7B4%7D%20%3D87" id="TexFormula1" title="x^{2} +\sqrt{4} =87" alt="x^{

Alja [10]

Answer:

<h2><em><u>ᎪꪀsωꫀᏒ</u></em></h2>

➪x= √ 85

Explanation:

x²+√4 = 87

=> x²+2 = 87

=> x² = 87-2

=> x²= 85

=> x= √85

7 0

3 years ago

Two carts on a straight track collide head on. The first cart was moving at 3.6 m/s in the positive x direction and the second w

zubka84 [21]

Answer:

vf₁ = -7.2 m/s : The first car moves in the negative x direction after the collision.

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁ = m kg : mass of the first car

m₂= 5m kg : mass of the second car

v₀₁ = 3.6 m/s : Initial velocity of m₁ , to the +x axis :

v₀₂= -2.4 m/s m/s : Initial velocity of m₂ , to the -x axis

vf₂= -0.24 m/s m/s : Final velocity of m₂ , to the -x axis

Problem development

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

We assume that the first car moves in the positive x direction after the collision., so, the sign of the final speeds is positive:

(m)*( 3.6) + (5m)*( -2.4) = (m)*vf₁ +(5m)*(- 0.24)

3.6m - 12m= m*vf₁ - 1.2m

We divided by m both sides of the equation

3.6 - 12= vf₁ -1.2

3.6 - 12 +1.2 = vf₁

-7.2 = vf₁

vf₁ = -7.2 m/s : The first car moves in the negative x direction after the collision.

4 0

3 years ago

A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th

Marysya12 [62]

Answer:

Mininmum acceleration required = $2.74 m/s^{2}$

Explanation:

Given,

Lift off speed = 125km/hr.

We know km/hr conversion factor to m/s is $\frac{5}{18}$ ,

Therefore,

Lift off speed required = $125 \times\frac{5}{18}$ m/s

Lift off speed required = 34.72222 m/s

Length of Takeoff Run given = 220 m.

So,

the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.

Consider the formula from kinematics,

$v^{2}-u^{2}=2as$ ,

where,

v - final velocity of particle,

u - initial velocity of particle

a - the constant acceleration at which the particle is moving with

s - distance travelled by particle

So at minimum acceleration,it reaches the takeoff speed at the end of the runway.

Therefore in the formula,

we use, v = Lift off speed = 34.72222 m/s;

u = 0 m/s (plane starts from rest);

a = minimum acceleration of the plane

s = 220 m;

Substituting these values in the formula, we get,

$(34.72222)^{2}-0 = 2\times a \times 220$

a = $\frac{(34.72222)^{2} }{2\times220}$

a = $2.74 m/s^{2}$ .

Therefore,the minimum acceleration of the plane required = $2.74 m/s^{2}$ .

7 0

3 years ago

Có một số điện trở giống nhau R0 = 3

Mama L [17]

Answer:

hlo

Explanation:

hlo olz mark me as brainlest

6 0

3 years ago