Which is the MOST accurate description of this organism?

A 1,492.3-kg airplane travels down the runway. Each of its four engines provides a force of

Fittoniya [83]

The acceleration of the air plane is $3.879 \mathrm{m} / \mathrm{s}^{2}$

<u>Explanation:</u>

Given:

The mass of the air plane = 1492.3 kg

Force of each four engine = 1447.5 N

So, the total force of four engines can be calculated as = 4(1447.5) = 5790 N

The force that acts on the object is equal to the product of mass (m) and its acceleration. It can express by the below formula,

$\text {Force }(F)=m \times \text { acceleration }(a)$

The above equation can be written as below to find acceleration,

$a=\frac{F}{m}$

Now. Substitute the given values, we get,

$a=\frac{5790}{1492.3}=3.879 \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago

8. When a 100 N bag of nails hangs motionless from a single vertical strand of rope, how many newtons of tension are exerted in

Svetllana [295]

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero. So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".

If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.

And if there are four ropes holding it up, then the sum of
the four tensions is 100N. If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.

8 0

3 years ago

A 30kg mass is brought from the earth to the moon what is the weight on the earth

ozzi

Answer:

Solution

verified

Verified by Toppr

Given:

Mass of body = 30 kg

gravitational acceleration on the moon = 1.62 m/s

2

Weight of the body on the moon = Mass of the body×gravitational acceleration on the moon=30×1.62=48 N

8 0

2 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago

¿Por qué los mensajes llegan en forma inmediata de un celular a otro?

KATRIN_1 [288]

Porque el internet ayuda a la rapidez del mensaje.

7 0

3 years ago