All categories

1 year ago

4. A plane accelerates down a runway due to the constant resultant force of

Physics

1 answer:

Doss [256]1 year ago

4 0

Answer:

See below

Explanation:

Net force acting on the plane after overcoming frictional forces is

2.7 x 10^6 - 6.8 x 10^4 - 8.5 x 10^5 = 1 782 000 N

Work = F x d

= 1 782 000 x 1400 m = 2.5 x 10^9 J or 2.5 x 10^6 kJ

You might be interested in

What goes through evaporation

Olegator [25]

Water goes through evaporation.

6 0

3 years ago

Read 2 more answers

What is The substance that dissolves the solute.

Ludmilka [50]

Answer:

Explanation:

Do you mean the solvent? If this is off the mark, let me know in a comment.

The solvent is something that the solute is (usually) soluble in.

3 0

2 years ago

What did solomon asch discover in his famous experiment on judging the lengths of lines?

BaLLatris [955]

Answer:

Asch (1956) found that group size influenced whether subjects conformed. The bigger the majority group (no of confederates), the more people conformed, but only up to a certain point.

Explanation:

4 0

2 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

2 years ago

Bob is pulling a 30 kg filing cabinet with a force of 200 N , but the filing cabinet refuses to move. The coefficient of static

Arada [10]

Answer:

<em>Magnitude of the Frictional force is 200 N</em>

Explanation:

The frictional force is the force that tries to oppose relative motion between two surfaces that are contacting. The coefficient of static friction is the coefficient of friction of a body that is not moving.

Newton's third law of motion states that action and reaction forces are equal and opposite. So the frictional force felt on the filing cabinet will be equal to the applied force pulling the cabinet.

Frictional force = Force applied

Force applied = 200 N

Therefore, the magnitude of the friction force on the filing cabinet is 200 N

5 0

3 years ago