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3 years ago

The acceleration due to gravity is ____.

Physics

2 answers:

skad [1K]3 years ago

8 0

The acceleration due to gravity is 9.89 m / s2.

<u>Explanation:</u>

Gravity was discovered by Newton when he was sitting under a tree and was watching an apple reaching the ground instead of going upwards. He fond that there is a gravitational pull attached to the ground of the earth which makes all the objects towards the ground. It is represented by 'g' and its value is equal to 9.8 m /s2.

ValentinkaMS [17]3 years ago

4 0

The acceleration which is gained by an object because of gravitational force is called its acceleration due to gravity. Its SI unit is m/s2. Acceleration due to gravity is a vector, which means it has both a magnitude and a direction. The acceleration due to gravity at the surface of Earth is represented by the letter g. It has a standard value defined as 9.80665 m/s2.[ However, the actual acceleration of a body in free fall varies with location. (Wikipedia)

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b) A satellite is in a circular orbit around the Earth at an altitude of 1600 km above the Earth's surface. Determine the orbita

asambeis [7]

Explanation:

The orbiting period of a satellite at a height h from earth' surface is

T=2πr32gR2

where r=R+h.

Then, T=2π(R+h)R(R+hg)−−−−−−−−√

Here, R=6400km,h=1600km=R/4

T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√

Putting the given values,

T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h

Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.

Let it be nR.Then

T=2π(R+nR)R(R+nRg)−−−−−−−−−−√

=2π(Rg)−−−−−√(1+n)32

=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32

(5075s)(1+n)32=(1.41h)(1+n)32

For T=24h, we have (24h)=(1.41h)(1+n)32

or (1+n)32=241.41=17

or 1+n(17)23=6.61

or n=5.61

The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.

3 0

3 years ago

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t

BabaBlast [244]

Answer:

$-\frac{8\pi}{3}rad/s^2$

Explanation:

To solve this problem we need to apply the concept related to Angular Acceleration. We can find it through the equation

$\omega_f^2-\omega_i^2=2\alpha\theta$

Where for definition,

$\omega_i = \frac{\theta}{t}$

The number of revolution $(\theta)$ was given by 20 times, then

$\omega_i = \frac{20*2pi}{5}$

$\omega = 8\pi rad/s$

We know as well that the salad rotates 6 more times, therefore in angle measurements that is

$\theta = 6*2\pi rad = 12\pi rad$

The cook at the end stop to spin, then using our first equation,

$0-8\pi = 2\alpha (12\pi)$

re-arrange to solve $\alpha$ ,

$\alpha = \frac{-8\pi}{2*12\pi}$

$\alpha = -\frac{8\pi}{3}rad/s^2$

We can know find the required time,

$\omega_f-\omega_i = \alpha t$

Re-arrange to find t, and considering that $\omega_f=0$

$t= \frac{\omega_i}{\alpha}$

$t=\frac{-8\pi}{-8\pi/3}$

$t=3s$

Therefore take for the salad spinner to come to rest is 3 seconds with acceleration of $-\frac{8\pi}{3}rad/s^2$

6 0

3 years ago

Read 2 more answers

The metric unit for weight force is _______ and the metric unit for force is ________.

Finger [1]

Answer:

Explanation:

Because i used google and it gave me both of the answers in like 5 seconds

7 0

3 years ago

a concave mirror forms a real image at 17 cm from the mirror surface along the principal axis. if the corresponding object is at

Step2247 [10]

Image distance (Di) = 17 cm
Object distance (Do) = 36 cm

Using mirror equation,

1/Di + 1/Do = 1/f
1/17 + 1/36 = 1/f
0.058824 + 0.027777 = 1/f
1/f = 0.086601
f= 11.5472 = 11.55 cm

8 0

3 years ago

If I start with 32g of Iodine-131, how much will I have after 24 days?

mihalych1998 [28]

Iodine-131. Iodine-131 decays by beta emission and has a half-life of 8.04 days. The principal gamma emission of 364 keV is considerably higher than the ideal for imaging with gamma cameras.

6 0

3 years ago