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3 years ago

Make the following conversion. 50.5 cm = _____ hm

Physics

2 answers:

NikAS [45]3 years ago

7 0

Answer: The value of 50.5 cm in hectometer is 0.00505

Explanation:

We are given 50.5 centimeters and we need to convert it into hectometers. So, for this we need to use the following conversion:

1 hectometer = 10000 centimeter

Now, to convert centimeters into hectometers, we divide the number by 10000, so we get:

$50.5cm=\frac{50.5}{10000}hm=0.00505hm$

Hence, the value of 50.5 cm in hectometer is 0.00505

Rina8888 [55]3 years ago

3 0

50.5cm= 0.00505hm Hope this helps!

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How many times did john glenn orbit the earth

galben [10]

Answer:

2 times

Explanation:

1st time on Feb. 20, 1962 at age 41

2nd time on Oct. 29, 1998 at age 77

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3 years ago

A helium balloon has an internal pressure of 2.5 atm when it occupies 4.5 L. If it was compressed until it had a pressure of 3.3

steposvetlana [31]

Answer: V = 3.4 L

Explanation: Use Boyle's Law to find the new volume. P1V1 = P2V2, derive for V2, then the formula will be V2= P1V1 / P2

V2 = 2.5 atm ( 4.5 L ) / 3.3 atm

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3 years ago

Read 2 more answers

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = $1.6.10^{-19}$ C

To determine work in joules, potential has to be in Volts, so:

$\Delta V=65.10^{-3}V$

Then, work is

$W=1.6.10^{-19}.65.10^{-3}$

$W=1.04.10^{-20}$

To move a potassium ion from the exterior to the interior of the cell, it is required $W=1.04.10^{-20}$ J of energy.

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3 years ago

The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in

Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly. We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

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3 years ago

Josh starts his sled at the top of a 2.9-m-high hill that has a constant slope of 25∘. After reaching the bottom, he slides acro

algol13

Answer:

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2(9.81)(2.9)

=56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v²- u² = 2 -μ g S

v=0 m/s

-(7.54)²=-2(0.06)(9.81)S

S=48.29 m

3 0

3 years ago