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Brrunno [24]
3 years ago
14

the distance of the earth to the sun is 1.5x10^11 m and one year is 365days if the distance to the sun increased six time the in

itial value how long it take the earth to orbit the sun?
Physics
1 answer:
I am Lyosha [343]3 years ago
6 0

Answer:

x = 2190[days]

Explanation:

This problem can be solved by means of a simple three rule:

1.5*10^{11}=365[days]\\ 6*(1.5*10^{11})=x\\\\x= \frac{6*(1.5*10^{11})*365[days]}{1.5*10^{11}} \\x=2190[days]

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suppose that a large cargo truck needs to cross a bridge. the truck is 30 m long and 3.2 m wide. the cargo exerts a force nof 54
Tamiku [17]
<span>No, because the truck applies more pressure than the bridge can support.</span>
7 0
3 years ago
Which three quantities can be used to calculate acceleration?
PtichkaEL [24]
D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
A and C are clearly incorrect, as mass and force (in terms of projectile motion) have no effect on an object's motion. B is incorrect because it is not useful to know the position or distance traveled, unless it will help you find displacement. Even then, you would not have enough information to use a kinematics equation to find a.
4 0
3 years ago
When the distance between two interacting objects doubles, the gravitational force is
Umnica [9.8K]

The gravitational force will be one quarter.

The gravitational force between two objects is given by the formula

F=GMm/r^2

here, r is the distance between the objects.

Thus the gravitational force is inversely proportional to the square of the distance between the objects, Therefore if the distance between two objects is doubled the force will be one quarter.

5 0
3 years ago
Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the s
sveta [45]

Answer:

b

Explanation:

Given:

- The ball is fired at a upward initial speed v_yi = 2*v

- The ball in first experiment was fired at upward initial speed v_yi = v

- The ball in first experiment was as at position behind cart = x_1

Find:

How far behind the cart will the ball land, compared to the distance in the original experiment?

Solution:

- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:

                                      v_yf = v_yi + a*t

Where, a = -9.81 m/s^2 acceleration due to gravity

            v_y,f = 0 m/s max height for both cases:

For experiment 1 case:

                                     0 = v - 9.81*t_1

                                      t_1 = v / 9.81

For experiment 2 case:

                                     0 = 2*v - 9.81*t_2

                                      t_2 = 2*v / 9.81

The total time for the journey is twice that of t for both cases:

For experiment 1 case:

                                     T_1 = 2*t_1

                                     T_1 = 2*v / 9.81

For experiment 2 case:

                                     T_2 = 2*t_2

                                     T_2 = 4*v / 9.81

- Now use 2nd equation of motion in horizontal direction for both cases:

                                     x = v_xi*T

For experiment 1 case:

                                     x_1 = v_x1*T_1

                                    x_1 = v_x1*2*v / 9.81

For experiment 2 case:

                                     x_2 =  v_x2*T_2

                                    x_2 = v_x2*4*v / 9.81

- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.

- Hence; take ratio of two distances x_1 and x_2:

                        x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v

Simplify:

                        x_1 / x_2 = 2  

- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.      

                                   

3 0
3 years ago
A man exerts a force of 180 N on a cart at an angle of 29° above the horizontal.
TiliK225 [7]
Answer:
This explanation from Physics/Geometry
60o |
|
| Fy (the vert. comp.)
↑
← 30o |
Fx (the horizontal componenet)

F = √(Fx2 + Fy2)
Fy = 50 · cos 60o = 50 · (1/2) = 25 N
Fx = 50 · cos 30o = 50 · (√3)/2 = 25√3 N
I see, that vector sign did not appear in my comment above, so the vector equation is
→ → →
F = 50 cos 30o i + 50 cos 60o j
Explanation:
7 0
3 years ago
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