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3 years ago

What is the magnitude of the orbital velocity of the earth in m/s?

1 answer:

8 0

Circumference C=2πr
C=2π(1.5x10^8)=9.42x10^8 

In 365 Days there are 8760hr 

V=distance/time 

V=(9.42x10^8)/8760=107534.2km/hr

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Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What

Kobotan [32]

Answer:

Mass of the wooden Block is 20g.

Explanation:

The buoyant force equation will be used here

Buoyant Force= ρ*g*1/2V Here density used is of water

m*g= ρ*g*1/2V

Simplifying the above equation

2m= ρ*V Eq-1

Also we know from the question that

ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )

Equating Eq-1 & Eq-2 we get

2m = m+0.020

m = 0.020kg

m = 20g

6 0

3 years ago

To calculate acceleration you must know both the objects velocity and_____

amm1812

You need to know the time as well.

8 0

3 years ago

Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d

dezoksy [38]

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

$W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J$

Hence, the required work done is 1786.17 J.

6 0

2 years ago

What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of $62.5 cm of H_2O$

i.e. $h=62.5 cm =0.625 m$

Effective area $A=51 cm^2$

initial Pressure $= 1 atm=101.325 kPa$

Gauge Pressure $P=\rho gh$

$\rho =density\ of\ water =1000 kg/m^3$

$P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa$

Force creates a pressure of $P_1$ which will be equal to Gauge Pressure

$P_1=\frac{F}{A}$

$P_1=P_{gauge}$

$\frac{F}{A}=5.937 kPa$

$F=5.937\times 51\times 10^{-4}\times 10^3$

$F=30.27 N$

6 0

4 years ago

What is the period of a wave with a frequency of 100 Hz and a wavelength of 2.0 m

spin [16.1K]

The answer for the following answer is answered below.

Therefore the time period of the wave is 0.01 seconds.
Therefore the option for the answer is "B".

Explanation:

Frequency (f):

The number of waves that pass a fixed place in a given amount of time.

The SI unit of frequency is Hertz (Hz)

Time period (T):

The time taken for one complete cycle of vibration to pass a given point.

The SI unit of time period is seconds (s)

Given:

frequency (f) = 100 Hz

wavelength (λ) = 2.0 m

To calculate:

Time period (T)

We know;

According to the formula;

f = $\frac{1}{T}$

Where,

f represents the frequency

T represents the time period

from the formula;

T = $\frac{1}{f}$

T = $\frac{1}{100}$

T = 0.01 seconds

Therefore the time period of the wave is 0.01 seconds.

5 0

3 years ago

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