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uranmaximum [27]

3 years ago

9

An airplane accelerates from rest to its required takeoff speed with an acceleration 5.0 m/s^2 after travelling on the runway a

distance of 0.75 km. Find the takeoff speed of the airplane.

1 answer:

Alex Ar [27]3 years ago

5 0

Answer:

Explanation:

.75 km = 750 m

acceleration a = 5 m/s²

distance s = 750 m

initial velocity u = 0

Final velocity v = ?

v² = u² + 2as

v² = 0 + 2 x 5 x 750

v = 86.60 m /s.

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When a steady direct current flows through a coil, the only opposition to the flow of current is the resistance of the wire from

Elanso [62]

Answer:

True

Explanation:

This is true statement because in DC circuit there is only resistance which comes into play of opposition of current.

In DC circuit there is no role of frequency in if there is no frequency then there will no inductive reactance and no capacitive reactance to oppose the current flow

So in DC circuit only resistance is responsible to oppose the flow of current

So this is a true statement

4 0

2 years ago

How come when an object has a greater mass, it has a greater Inertia? and the other way around.

Vedmedyk [2.9K]

Inertia is a property of a body to resist the change in linear state of motion. It is measured by the mass of the body. An object having greater mass performing linear motion is difficult to stop. Hence greater the mass greater is its inertia. Hope it helps.

4 0

2 years ago

What is the electric potential 15 cm above the center of a uniform charge density disk of total charge 10 nC and radius 20 cm?

alisha [4.7K]

Answer:

b) 450 V

Explanation:

We are given that

Total charge, q=10nC= $10\times 10^{-9} C$

$1nC=10^{-9}C$

Radius, r=20 cm= $\frac{20}{100}=0.2m$

1 m=100 cm

x=15 cm=0.15 cm

We have to find the electrical potential 15 cm above the center of a uniform charge density disk .

We know that

$\sigma=\frac{q}{A}=\frac{q}{\pi r^2}$

$\sigma=\frac{10\times10^{-9}}{3.14\times (0.2)^2}$

Where $\pi=3.14$

$\sigma=7.96\times 10^{-8}C/m^2$

Electric potential, $V=\frac{\sigma}{2\epsilon_0}(\sqrt{x^2+r^2}-x)$

Where $\epsilon_0=8.85\times 10^{-12}$

Using the formula

$V=\frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}(\sqrt{(0.15)^2+(0.2)^2}-0.15)$

$V=449.7 V\approx 450V$

Hence, option b is correct.

7 0

2 years ago

Read 2 more answers

A car drives 500m in 2 minutes what is the cars speed?

love history [14]

250 mph? i think that’s correct lol, if it is mark brainliest please;)

6 0

2 years ago

Read 2 more answers

A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?

Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

$H_m = \frac{u^2sin^2 \theta}{2g}$

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

$H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta = sin^{-1}(0.2123)\\\\\theta = 12.26^0$

Therefore, the angle of projection is 12.26⁰.

6 0

2 years ago