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3 years ago

6

A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s

acceleration compared to the gravitational field at the location of the satellite?

2 answers:

bulgar [2K]3 years ago

4 0

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

$E = \frac{F}{m}$

now in order to find the acceleration of the satellite we know by Newton's II law

$F = ma$

so we will have

$a = \frac{F}{m}$

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

PSYCHO15rus [73]3 years ago

4 0

Answer:

The gravitational field and the acceleration point in the same direction.

The magnitudes of the acceleration and the gravitational field strength are equal.

These are the two answers

Explanation:

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n200080 [17]

Answer:

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Explanation:

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3 years ago

I NEED MAJOR HELP ON THIS AS WELL PLEASE SOMEONE HELP ME

yuradex [85]

Answer:

The total distance is 130.2 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the cart starts from rest, i.e. its initial speed is zero.

$v_{f} =v_{o} +(a*t)$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 3 [m/s²]

t = time = 8[s]

Vf = 0 + (3*8)

Vf = 24 [m/s]

With this velocity we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

24² = 0 + (2*3*x)

x = 576/(6)

x = 96 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

$v_{f}= v_{i} -(a*t)$

Vf = final velocity [m/s]

Vi = initial velocity = 24 [m/s]

a = desacceleration = 1.6 [m/s²]

t = time = 15 [s]

Vf = 24 - (1.6*15)

Vf = 21.6 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} -2*a*x$

where

x = distance traveled [m]

21.6² = 24² - (2*1.6*x)

x = 109.44/(3.2)

x = 34.2 [m]

Note: The negative sign in the equations is because the car is desaccelerating, it means its velocity is decreasing.

Therefore the total distance is Xt = 34.2 + 96 = 130.2 [m].

5 0

3 years ago

Ground reaction force acting on carter

mezya [45]

We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.

I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.

8 0

3 years ago

A pitcher throws a baseball that reaches the catcher in 0.75 s. The ball curves because it is spinning at an average angular vel

7nadin3 [17]

The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as

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

Here,

$\theta$ = Angular displacement

t = time

The angular velocity is given as

$\omega = 230rev/min$

PART A) The angular velocity in SI Units will be,

$\omega = 230rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega = \frac{23}{3}\pi rad/s \approx 24.08rad/s$

PART B) From our first equation we can rearrange to find the angular displacement then

$\theta = \omega t$

Replacing,

$\theta = (24.08)(0.75)$

$\theta = 18.06 rad$

4 0

3 years ago

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr

Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

$I(t)=0.88e^{-t/6\times3600s}$

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

$q=\int\limits {I} \, dt$

Here the limits of integration is from 0 to infinite. So,

$q=\int\limits {0.88e^{-t/6\times3600s}}\, dt$

$q=0.88\times(-6\times3600)(0-1)$

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

$N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}$

N = 1.2 x 10²³

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3 years ago