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2 years ago

Two objects of masses m1 = 0.56 kg and m2 = 0.88 kg are placed on a horizontal

Physics

1 answer:

ziro4ka [17]2 years ago

3 0

Explanation:

Stern et al. (1999) and Stern (2000), define this variable as those general visions about the world, reflected in the beliefs that people express about their relationship with the environment and nature.हेलो फ्रेंड्स मारो किसी को इनबॉक्स कैसे करें

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The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e

djverab [1.8K]

The gravitational field is the Force divided by the mass

Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.

g = F / m

g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg

Note that the unit N/kg is equivalent to m/s^2

8 0

3 years ago

Compare the maximum rate of heat transfer to the basal metabolic rate by converting a bmr of 88 kcal/hr into watts. what is the

elena-14-01-66 [18.8K]

Explanation :

It is given that,

BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :

We know that, 1 kilocalorie = 4184 joules

So, $1\ kcal/h=\dfrac{1\times 4184\ J}{3600\ sec}$

$1\ kcal/h=1.16\ J/sec$

J/sec is nothing but watts.

So, $1\ kcal/h=1.16\ watts$

and $88\ kcal/h=88\times 1.16\ watts = 102.08\ watts$

So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.

8 0

3 years ago

Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha

ikadub [295]

We have that the spring constant is mathematically given as

$k=2.37*10^{11}N/m$

Generally, the equation for angular velocity is mathematically given by

$\omega=\sqrt{k}{m}$

Where

k=spring constant

And

$\omega =\frac{2\pi}{T}$

Therefore

$\frac{2\pi}{T}=\sqrt{k}{n}$

Hence giving spring constant k

$k=m((\frac{2 \pi}{T})^2$

Generally

Mass of earth $m=5.97*10^{24}$

Period for on complete resolution of Earth around the Sun

$T=365 days$

$T=365*24*3600$

Therefore

$k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2$

$k=2.37*10^{11}N/m$

In conclusion

The effective spring constant of this simple harmonic motion is

$k=2.37*10^{11}N/m$

For more information on this visit

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8 0

3 years ago

An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination o

Ann [662]

Answer:

θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

v_fly² = v_nort² + v_air²

v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

sin θ = v_air / v_fly

θ = sin⁻¹ (v_air / v_fly)

let's calculate

θ = sin⁻¹ (10/120)

θ = 4.78º

with respect to the vertical or 4.78 to the north-east

5 0

3 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

6 0

3 years ago

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