When 2 forces are applied in opposite directions, how do you calculate the net force?

A plane lands on a runway with a speed of 115 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2

marin [14]

Answer:

 The planes average acceleration in magnitude and direction = 8.846 m/s² moving east

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I Unit of acceleration is m/s². Acceleration is a vector quantity because it can be represented both in magnitude and in direction.

Acceleration can be represented mathematically as

a = v/t.................................... Equation 1

Where a = acceleration, v = velocity, t= time.

Given: v = 115 m/s, t = 13.0 s

Substituting these values into equation 1

a = 115/13

a = 8.846 m/s² moving east

Thus the planes average acceleration in magnitude and direction = 8.846 m/s² moving east

4 0

2 years ago

Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

1)

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

2)

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

3)

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

4)

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0

2 years ago

Nvm i got the answer but free points

faust18 [17]

Answer:

thank you 谢谢

Explanation:

8 0

3 years ago

Read 2 more answers

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

2 years ago

A certain heat engine takes in 300 J of energy from a hot source and then transfers 200 J of that energy to a colder object. Wha

Greeley [361]

Answer:

The efficiency is 0.33, or 33%.

Explanation:

From the thermodynamics equations, we know that the formula for the efficiency of a heat engine is:

$\eta=1-\frac{Q_2}{Q_1}$

Where η is the efficiency of the engine, Q_1 is the heat energy taken from the hot source and Q_2 is the heat energy given to the cold object. So, plugging the given values in the formula, we obtain:

$\eta=1-\frac{200J}{300J}=0.33$

This means that the efficiency of the heat engine is 0.33, or 33% (The efficiency of an engine is dimensionless).

5 0

2 years ago