Ask question

Ask question

All categories

3 years ago

9

A hovercraft is being driven across a lake on a very windy day. The wind exerts a force of 5000 N north on the hovercraft. The p

ropellers driving the hovercraft also exert a force.
What is the net force (to 3.s.f ) on the hovercraft when the force applied to the hovercraft by the propellers is: 10 000 N west

1 answer:

krok68 [10]3 years ago

4 0

Answer:

The net force on the hovercraft is 11200 N.

Explanation:

Given;

force exerted on the hovercraft by wind, F₁ = 5000 N north

force exerted on the hovercraft by the propeller, F₂ = 10,000 N west

The net force on the hovercraft is calculated as;

$F_{net} = \sqrt{F_1^2 + F_2^2} \\\\F_{net} = \sqrt{5000^2 + 10,000^2} \\\\F_{net} = 11180.34 \ N\\\\F_{net} = 11200 \ N \ (3.s.f)$

Therefore, the net force on the hovercraft is 11200 N.

You might be interested in

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

3 0

3 years ago

These two cubes represent a small cell and large cell .

Elena L [17]

Answer:

Idk

Explanation:

7 0

3 years ago

he work function of a certain metal is 1.90 eV. What is the longest wavelength of light that can cause photoelectron emission fr

bixtya [17]

Answer:

6538.8 Angstrom

Explanation:

work function, w = 1.9 eV = 1.9 x 1.6 x 10^-19 J = 3.04 x 10^-19 J

Let the longest wavelength is λ.

W = h c / λ

λ = h c / W

λ = (6.626 x 10^-34 x 3 x 10^8) / (3.04 x 10^-19)

λ = 6.5388 x 10^-7 m = 6538.8 Angstrom

Thus, the longest wavelength is 6538.8 Angstrom.

3 0

3 years ago

5 examples of neutons 3rd law of motion

Lemur [1.5K]

Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

3 0

3 years ago

Read 2 more answers