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2 years ago

An airplane flies with a velocity of 750. kilometers

Physics

2 answers:

ra1l [238]2 years ago

6 0

B. 650 km/h is the correct answer

viktelen [127]2 years ago

5 0

Answer:

650 km/hr

Explanation:

Draw a right triangle from (0.0) (Point A) down 30 degrees and to the right for a length of 750 (Point B). Then draw a line from B up to the x axis to make a right angle (Point C). Use the cosine function to find line AC, the vector portion of AB that lies of the x (East) axis. Cosine(30)= Adjacent/Hypotenuse.

Cos(30) = AC/750

750*(cos(30)) = AC

AC = 649.5 km/hr

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If a net force of 250 N causes an object to accelerate at 20m/s^2 what must its mass be?

aleksandr82 [10.1K]

The answer is 12.5 kg because 250N / 20m/s^2

I hope that helped

7 0

2 years ago

A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³

Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

= 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

9.5 N = 10 N - weight of water displaced

Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

= 0.051 kg

Therefore the volume of the water displaced,

V = m / ρ

= 0.051 / 1000

= 5.1 x 10⁻⁵ m³

<em>The volume of water displaced is equal to the volume of alcohol displaced</em>

Therefore, the mass of the alcohol displaced, = V x ρ

= 5.1 x 10⁻⁵ x 800

= 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

= 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

W₀ = W - w

= 10 N - 0.4 N

= 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0

2 years ago

ANSWER ASAP PLZZZZ!!!!!!!!

igor_vitrenko [27]

Answer: your correct answer is a i took the test

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2 years ago

The phases of the moon depend on how much of the lighted side of the moon can be seen from Earth.

Ostrovityanka [42]

Answer:

true

Explanation:

if there is no light it's different from when there is

3 0

2 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

2 years ago