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2 years ago

How are feathers heavy as steel?

1 answer:

6 0

Answer: This is because they both weigh a pound. As the density of steel is much greater than the density of feathers, a pound of feathers would make up a much greater volume.

Explanation: brainliest plz :)

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If a box is pulled with a force of 100 N at an angle of 25

love history [14]

The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

<u>Given the following data:</u>

Force = 100 Newton.

Angle of inclination = 25°

To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

$F_x = Fcos \theta\\\\F_x =100 \times cos25\\\\F_x =100 \times 0.9063$

Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

$F_y = Fsin \theta\\\\F_y =100 \times sin25\\\\F_y =100 \times 0.4226$

Fy = 42.26 Newton.

Read more on horizontal component here: brainly.com/question/4080400

6 0

2 years ago

Which of the following results when a crest and trough meet?

kifflom [539]

D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.

6 0

2 years ago

A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro

ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let $u_1, u_2 , v_f$ be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

$m=2.50\times 10^{4}$

Conserving momentum

$mu_1+3mu_2=4mv_f$

$4m+6m=4mv_f$

$v_f=2.5 m/s$

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

$Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J$

$Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J$

$Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ$

4 0

3 years ago

. A bathysphere used for deep-sea exploration has a radius of 1.50 m and a mass of 1.20 104 kg. To dive, this submarine takes on

dangina [55]

Answer:

Explanation:

Given that,

Bathysphere radius

r = 1.5m

Mass of bathysphere

M = 1.2 × 10⁴ kg

Constant speed of descending.

v = 1.2m/s

Resistive force

Fr = 1100N upward direction

Density of water

ρ = 1.03 × 10³kg/m³

The volume of the bathysphere can be calculated using

V = 4πr³ / 3

V = 4π × 1.5³ / 3

V = 14.14 m³

The Bouyant force can be calculated using

Fb = ρgV

Fb = 1.03 × 10³ × 9.81 × 14.14

Fb = 142,846.18 N

Buoyant force is acting upward

Weight of the bathysphere

W = mg

W = 1.2 × 10⁴ × 9.81

W = 117,720 N

Weight is acting downward

The net positive buoyant using resolving

Fb+ = Fb — W

Fb+ = 142,846.18 — 117,720

Fb+ = 25,126.18 N

The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force

W = Fb+ + Fr

W = 25,126.18 + 1100

W = 26,226.18

mg = 26,226.18

m = 26,226.18 / 9.81

m = 2673.4kg

Mass of submarine is 2673.4kg

8 0

2 years ago

(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t

ratelena [41]

Answer:

radius r is 0.414 R

Explanation:

Given data

FCC octahedral site

atomic radius R

to find out

radius r

solution

we know that at FCC octahedral

length of side = 2 R + 2r

and by pythagorean theorem

a = 2√2R

here a = 2R + 2r

so 2R + 2r = 2√2R

so r = ( √2R )- R

r = 0.414 R

so radius r is 0.414 R

5 0

3 years ago