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marshall27 [118]

3 years ago

7

The standard kilogram is a platinum-iridium cylinder 39.0 mm in height and 39.0 mm in diameter. What is the density of the mater

ial?

1 answer:

KengaRu [80]3 years ago

7 0

Density is mass per unit of volume, usually measured in g/cm³. So first determine the volume of the cylinder:

<em>v</em> = <em>π</em> <em>r </em>² <em>h</em> = <em>π</em> ((39.0 mm) / 2)² (39.0 mm)

<em>v</em> = 59,319/4 <em>π</em> mm³ ≈ 46,589 mm³ ≈ 46.589 cm³

Then the density is

<em>ρ</em> = <em>m</em> / <em>v</em>

<em>ρ</em> = (1 kg) / <em>v</em> = (1000 g) / <em>v</em>

<em>ρ</em> ≈ 21.464 g/cm³

Or, if you want to preserve the given units, that's

<em>ρ</em> ≈ 0.00002146 kg/mm³

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The compass of an airplane indicates that it is headed due north and its airspeed indicator shows that it is moving through the

yarga [219]

Answer:

Airplane speed relative to the ground is 260 km/h and θ = 22.6º direction from north to east

Explanation:

This is a problem of vector composition, a very practical method is to decompose the vectors with respect to an xy reference system, perform the sum of each component and then with the Pythagorean theorem and trigonometry find the result.

Let's take the north direction with the Y axis and the east direction as the X axis

Vy = 240 km / h airplane

Vx = 100 Km / h wind

a) See the annex

Analytical calculation of the magnitude of the speed and direction of the aircraft

V² = Vx² + Vy²

V = √ (240² + 100²)

V = 260 km/h

Airplane speed relative to the ground is 260 km/h

Tan θ = Vy / Vx

tan θ = 100/240

θ = 22.6º

Direction from north to eastb

b) What direction should the pilot have so that the resulting northbound

Vo = 240 km/h airplane

Vox = Vo cos θ

Voy = Vo sin θ

Vx = 100 km / h wind

To travel north the speeds the x axis (East) must add zero

Vx -Vox = 0

Vx = Vox = Vo cos θ

100 = 240 cos θ

θ = cos⁻¹ (100/240)

θ = 65.7º

North to West Direction

The speed in that case would be

V² = Vx² + Vy²

To go north we must find Vy

Vy² = V² - Vx²

Vy = √( 240² - 100²)

Vy = 218.2 km / h

8 0

4 years ago

What is the acceleration of a car with the mass of 1,980 kg with

Nikolay [14]

The acceleration of a car with the mass of 1,980kg with a net horizontal force of 1,000 N is 0.51m/s².

<h3>HOW TO CALCULATE ACCELERATION?</h3>

The acceleration of a body can be calculated using the following formula:

a = F ÷ m

Where;

a = acceleration (m/s²)
F = Force (N)
m = mass (kg)

According to this question, a car has a mass of 1980kg and horizontal force of 1000N. The acceleration is as follows:

a = 1000/1980

a= 0.51m/s²

Therefore, the acceleration of a car with the mass of 1,980kg with a net horizontal force of 1,000 N is 0.51m/s².

Learn more about acceleration at: brainly.com/question/12134554

3 0

2 years ago

When a pitcher throws a baseball, it reaches a top speed of 39 m/s. If the baseball takes 1.5 seconds to travel from the pitcher

BARSIC [14]

Answer:

a.26

Explanation:

39 divide by 1.5=26

7 0

2 years ago

Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux

Fofino [41]

Answer:

a

$\phi = 1.78 *10^{-7} \ Weber$

b

$L = 1.183 *10^{-7} \ H$

Explanation:

From the question we are told that

The radius is $r = 6 \ cm = \frac{6}{100} = 0.06 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic flux of the coil is mathematically represented as

$\phi = B * A$

Where B is the magnetic field which is mathematically represented as

$B = \frac{\mu_o * I}{2 * r}$

Where $\mu_o$ is the magnetic field with a constant value $\mu_o = 4\pi * 10^{-7} N/A^2$

substituting value

$B = \frac{4\pi * 10^{-7} * 1.50 }{2 * 0.06}$

$B = 1.571 *10^{-5} \ T$

The area A is mathematically evaluated as

$A = \pi r ^2$

substituting values

$A = 3.142 * (0.06)^2$

$A = 0.0113 m^2$

the magnetic flux is mathematically evaluated as

$\phi = 1.571 *10^{-5} * 0.0113$

$\phi = 1.78 *10^{-7} \ Weber$

The self-inductance is evaluated as

$L = \frac{\phi }{I}$

substituting values

$L = \frac{1.78 *10^{-7} }{1.50 }$

$L = 1.183 *10^{-7} \ H$

7 0

3 years ago

A ball is thrown from the top of a building with an initial velocity of 21.9 m/s straight upward, at an initial height of 51.6 m

nalin [4]

Part A)

when ball will reach to highest point then it's speed will become zero

so we can use kinematics to find the time

$v_f = v_i + at$

$0 = 21.9 + (-9.8) t$

$0 = 21.9 - 9.8 t$

$t = 2.23 s$

Part b)

for finding the maximum height we can use another kinematics equation

$v_f^2 - v_i^2 = 2ad$

$0 - 21.9^2 = 2(-9.8)(H)$

$H = \frac{21.9^2}{19.6} = 1.12 m$

so it will rise to 1.12 m from the point of projection

Part C)

Ball will take double the time which it take to reach the top point.

So here the time to reach the top is 2.23 s

so time taken by the ball to reach at same point after projection is given as

$t = 2(2.23) = 4.46 s$

Since ball have reached to same point so the final velocity must be same as initial velocity

so we have

$v_f = 21.9 m/s$ downwards

Part d)

when ball reached to the bottom

displacement of ball = -51.6 m

$a = -9.8 m/s^2$

$v_i = 21.9 m/s$

now by kinematics we have

$d = v_i t + \frac{1}{2}at^2$

$-51.6 = (21.9)t + \frac{1}{2}(-9.8)t^2$

$4.9 t^2 - 21.9 t - 51.6 = 0$

by solving above equation we have

$t = 6.2 s$

now for the velocity at that instant we have

$v_f = v_i + at$

$v_f = 21.9 - (9.8) (6.2)$

$v_f = -38.6 m/s$

so its velocity is 38.6 m/s downwards

Part e)

for the position of ball at t = 5.35 s we can use

$d = v_i t + \frac{1}{2}at^2$

$d = 21.9(5.35) + \frac{1}{2}(-9.8)(5.35)^2$

$d = -23.1 m$

so it is 23.1 m below the initial position from which it is thrown

now for the velocity we can say

$v_f = v_i + at$

$v_f = 21.9 + (-9.8)(5.35)$

$v_f = -30.53 m/s$

so it will be moving downwards with speed 30.53 m/s

8 0

3 years ago