Will give brainlist!! Please help!!

A jet airliner moving initially at 693 mph (with respect to the ground) to the east moves into a region where the wind is blowin

Nesterboy [21]

Answer:

The new speed of the aircraft with respect to the ground is 1414.3 mph.

Explanation:

Given that,

Angle = 37°

Velocity of jet airliner = 693 mph

Velocity of wind = 798 mph

We know that,

The new velocity of the aircraft with respect to the ground

$v=v_{a}+v_{w}$

We need to calculate the new speed of the aircraft with respect to the ground

Using formula for velocity

$v=\sqrt{(v_{a})^2+(v_{w})^2+2v_{a}v_{w}\cos\theta}$

Put the value into the formula

$v=\sqrt{(693)^2+(798)^2+2\times693\times798\cos37}$

$v=1414.3\ mph$

Hence, The new speed of the aircraft with respect to the ground is 1414.3 mph.

7 0

3 years ago

A neuron that is activated when a mosquito lands on your arm

mash [69]

<span> The spinal neurons involved in the tingling sensation caused by a light touch are different from those transmitting pain or a 'chemical' itch, the latter elicited by a mosquito bite or a skin wound that is healing.
hope this helps</span>

8 0

3 years ago

A loop circuit has a resistance of R1 and a current of 1.8 A. The current is reduced to 1.6 A when an additional 3.8 Ω resistor

Allisa [31]

Answer:

Value of $R_1=30.4ohm$

Explanation:

We have given

In first case resistance is $R_1$ and current is 1.8 A

Let the potential difference is v

So $1.8=\frac{v}{R_1}$ ----eqn 1

In second case resistance is $R_1+3.8$ and current is 1.6 A and potential difference will be as it is a series connection

So $1.6=\frac{v}{R+3.8}$ ----eqn 2

From eqn 1 and eqn 2

$1.8R_1=1.6R_1+6.08$

$R_1=30.4ohm$

6 0

3 years ago

Read 2 more answers

Need help with number 5 ASAP please

topjm [15]

Explanation:

The current flow of electrons for all of the light bulbs would be interrupted. It is similar or 'breaking' a circuit.

5 0

3 years ago

In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable

pav-90 [236]

Answer:

Approximately $3.98\; \rm m \cdot s^{-2}$ .

Assumption: air resistance on the rocket is negligible. Take $g = \rm 9.81\; m \cdot s^{-2}$ .

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

$\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ .

Note that in this case, the uppercase letter $\rm M$ in the units stands for "mega-", which is the same as $10^6$ times the unit that follows. For example, $\rm 1\; Mg = 10^6\; g$ , while $\rm 1\; MN = 10^6\; N$ .

Convert the mass of the rocket and the thrust of its engines to SI standard units:

The standard unit for mass is kilograms: $\displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g} = 552 \times 10^3 \; kg$ .
The standard for forces (including thrust) is Newtons: $\text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N$ .

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

Thrust (which is supposed to go upwards), and
Weight (downwards due to gravity.)

The thrust on the rocket is already known to be $\rm 7.61 \times 10^6\; N$ . Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately $9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}$ . Hence, the weight on the rocket would be approximately $9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N$ .

The magnitude of the net force on the rocket would be

$\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}$ .

Apply the formula $\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

$\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg} \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}$ .

6 0

3 years ago