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alex41 [277]
2 years ago
15

Does heavy force work under conditions of weightlessness in such an artificial earth satellite?

Physics
1 answer:
Afina-wow [57]2 years ago
3 0

Answer: That is to say, a satellite is an object upon which the only force is gravity. Once launched into orbit, the only force governing the motion of a satellite is the force of gravity. Newton was the first to theorize that a projectile launched with sufficient speed would actually orbit the earth.

Explanation: Because Earth-orbiting objects follow elliptical paths around Earth and not a straight line, forces cannot, by definition, be balanced. Force is directional. It is a push or a pull in a particular direction.

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You are walking in Paris alongside the Eiffel Tower and suddenly a croissant falls on your head and knocks you to the ground. If
Alexxandr [17]

Answer:

7,79 seconds

Explanation:

{\displaystyle {\overline {a}}={\frac {\Delta v}{t}}}

You need to use the acceleration formula. A is acceliration, \displaystyle \Delta \mathbf {v} is change in velocity and t is time.

You  need to multiply the formula with t and divide by a and you get

a*t=\displaystyle \Delta \mathbf {v}

t= \displaystyle \Delta \mathbf {v}/a

after that you just need to insert the numbers

change in velocity is 76.4 minus 0.

acceliration is gravitational acceleration which is 9.81.

After that you get

t=76.4/9.81

t= 7,787971458 s

6 0
3 years ago
Read 2 more answers
2. A portion of the body receives 0.15 mGy from radiation with a quality factor Q = 6 and 0.22 mGy from radiation with Q = 10. (
DiKsa [7]

Answer with Explanation:

We are given that

D_1=0.15mGy

D_2=0.22mGy

Q_1=6

Q_2=10

a.We have to find the total dose

Total dose=D=D_1+D_2

Using the formula then, we get

D=0.15+0.22

D=0.37mGy

b.We have to find the total dose equivalent

Total dose equivalent=H=D_1Q_1+D_2Q_2

Using the formula

H=0.15(6)+0.22(10)

H=3.1mSv

7 0
2 years ago
A flatbed truck is carrying an 800-kg load of timber that is not tied down. The maximum friction force between the truck bed and
ycow [4]

Answer:

Acceleration a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

Explanation:

For the truck to accelerate without losing its load.

Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.

Fa ≤ F(friction)

But;

Fa = mass × acceleration

Fa = ma

ma ≤ F(friction)

a ≤ (F(friction))/m ......1

Given;

Fa = mass × acceleration

Fa = ma

mass m = 800 kg

F(friction) = 2400 N

Substituting the given values into equation 1;

a ≤ F(friction)/m

a ≤ 2400N/800kg

a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

4 0
3 years ago
An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s.
Veseljchak [2.6K]

Answer:

a) -1.25 rev/s² and 23.3 rev

b)  2.67s

Explanation:

a) ωФ_o_z = (500 rev/min)(1min/ 60s) => 8.333 rev/s

ωФ_Z= (200 rev/min)(1min/ 60s) => 3.333rev/s

time 't'= 4 s

angular acceleration 'αФ_Z'=?

constant angular acceleration equation is given by,

ωФ_Z= ωФ_o_z + αФ_Zt

αФ_Z= (ωФ_Z - ωФ_o_z )/t => (3.333-8.333)/4

αФ_Z= -1.25 rev/s²

θ-θФ_o = ωФ_o_z t + 1/2αФ_Zt²

      =(8.333)(4) + 1/2 (-1.25)(4)²

      =23.3 rev

b) ωФ_Z=0   (comes to rest)

ωФ_o_z = 3.333 rev/s

αФ_Z= -1.25 rev/s²

ωФ_Z= ωФ_o_z + αФ_Zt

t= (ωФ_Z - ωФ_o_z)/αФ_Z => (0- 3.333)/-1.25

t= 2.67s

3 0
2 years ago
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
uranmaximum [27]

Answer:

fast. i have spoken

Explanation:

3 0
3 years ago
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