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umka2103 [35]
3 years ago
10

An object 1,000 km above earths surface would accelerate

Physics
1 answer:
Verizon [17]3 years ago
5 0

The acceleration which is gained by an object because of the 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 formula is ‘the change in velocity= gravity x time’ The acceleration due to gravity at the surface of Earth is represented as g. It has a standard value defined as 9.80665 m/s2.[1]

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A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
3 years ago
Question 3. AP PHYSICS. Currently learning about torque so I assume you apply torque. I don't get it at all
8090 [49]

wow that is confusing


8 0
3 years ago
Determine the torque<br> produced by a perpendicular force of 75<br> N at the end of a 0.2 m wrench.
trasher [3.6K]

Answer:c

Explanation:

Because I took uty

7 0
2 years ago
Read 2 more answers
A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m
Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, I_1=900\ kg.m^2

Initial angular velocity of the platform, \omega=0.95\ rad/s

Part A,

Let \omega_2 is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

I_1\omega_1=I_2\omega_2

Here, I_2=I_1+mr^2

I_1\omega_1=(I_1+mr^2)\omega_2

900\times 0.95=(900+70\times (2.9)^2)\omega_2

Solving the above equation, we get the value as :

\omega_2=0.574\ rad/s

Part B,

The initial rotational kinetic energy is given by :

k_i=\dfrac{1}{2}I_1\omega_1^2

k_i=\dfrac{1}{2}\times 900\times (0.95)^2

k_i=406.12\ rad/s

The final rotational kinetic energy is given by :

k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2

k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2

k_f=245.24\ rad/s

Hence, this is the required solution.

5 0
3 years ago
Use the drop-down menus to complete the sentences.
Grace [21]

Answer:

1) thinner, diverge

2) thicker, converge

Explanation:

i got it right

5 0
2 years ago
Read 2 more answers
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