6+8t+3t²=0<br> find time

(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance

Komok [63]

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity F=W= mg [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

=132 kg ×9.8 m/s^2

=1293.6 kg m/s^2

=1293.6 N [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

$=\frac{1}{4} *1293.6N$

$=323.4 N$

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

$F_{net} =1293.6N -323.4N$

$=970.2 N$

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e

$F_{net} =ma$ [Here a is the acceleration]

$a =\frac{F_{net} }{m}$

$= \frac{970.2}{132} m/s^2$

$=7.35 m/s^2$

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

we know that F= ma

=m×0

=0 N

Hence they will not get any force when they will descend with a uniform speed.

4 0

2 years ago

A wave is moving at 18 m/s. If its wavelength is 3 meters, what is its frequency?

Vlada [557]

here's the solution,

we know that,

=》

$wave \: speed = wavelength \times frequency$

so,

=》

$18 = 3 \times f$

=》

$f = \dfrac{18}{3}$

=》

$f = 6$

frequency = 6 hertz

7 0

3 years ago

Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.

Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

$L = L_0 \sqrt{1-\frac{v^2}{c^2} }$

L = 1 light year

L₀ = 100,000 light year

$1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }$

$1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }$

$\frac{1}{100,000} = \sqrt{1-\frac{v^2}{c^2} }$

$v = 0.999999995$ $c$

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

brainly.com/question/17423632

#SPJ4

5 0

1 year ago

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph

arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

$I_{CM} = \dfrac{2}{5}mR^2$ .....(1)

now, moment of inertia about tangent

$I= \dfrac{2}{5}mR^2 + mR^2$

$I= \dfrac{7}{5}mR^2$ ...........(2)

dividing equation (1)/(2)

$\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}$

$\dfrac{I_{CM}}{I}=\dfrac{2}{7}$

$I_{CM}=\dfrac{2}{7}I$

the correct answer is option E

4 0

3 years ago

The magnetic force on a straight wire 0.69 m long is 1.5 x 10-3 N. The current in the wire is 16.9 A. What is the magnitude of t

VikaD [51]

Answer:

Magnitude of magnetic field is 1.29 x 10⁻⁴ T

Explanation:

Given :

Current flowing through the wire, I = 16.9 A

Length of wire. L = 0.69 m

Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N

Consider B be the applied magnetic field.

The relation to determine the magnetic force experienced by current carrying wire is:

F = ILBsinθ

Here θ is the angle between magnetic field and current carrying wire.

According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:

F = ILB

$B=\frac{F}{IL}$

Substitute the suitable values in the above equation.

$B=\frac{1.5\times10^{-3} }{16.9\times0.69}$

B = 1.29 x 10⁻⁴ T

7 0

3 years ago

6+8t+3t²=0 find time​

6+8t+3t²=0 find time