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Vaselesa [24]
2 years ago
6

A 2.34 kg book is dropped from a height of +3.1 m. (a) What is its acceleration?

Physics
1 answer:
frosja888 [35]2 years ago
8 0
<span> 9.8 m/s^2 is the answer</span>
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Https://youtu.be/F_DEsMO_Wq8

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3 0
3 years ago
An airplane is accelerating forward due to a force of 3320 newtons acting
Darya [45]

Two opposite forces generate a force whose magnitude is the difference of the two magnitudes, and whose direction is the same of the strongest one.

We have 3320 newtons forward, and 550 backwards.

So, the net force is forward, and has magnitude 3320-550=2770

5 0
3 years ago
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Explanation:

5 0
3 years ago
Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe
nignag [31]

Answer:

a) x = 40 t , y = 39 t ,  z = 6 + 32 t - 16 t ²,   b)     x = 80 feet ,  y = 78 feet , the ball came into the field  

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

          x₀ = 0

          y₀ = 0

          z₀ = 6 feet

x-axis (towards end zone,  GOAL zone)

         x = xo + v₀ₓ t

        x = 40 t

y-axis (field width)

        y = y₀ + v_{oy} t

        y = 39 t

z axis (vertical)

        z = z₀ + v_{oz} t - ½ g t²

        z = 6 + 32 t - ½ 32 t²

        z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

          z = 6

          6 = 6 + 32 t - 16 t²

          (t - 2)t  = 0

           t=0 s

           t= 2 s

The ball position

          x = 40 2

          x = 80 feet

           

          y = 39 2

          y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

             x_total = 150 feet

             y _total = 80 feet

so we can see that the ball came into the field

6 0
3 years ago
An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid
serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is r_x=140 km

Distance of astronaut From asteroid Y is r_y=581 km

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

F_x=F_y

\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}

cancel out the common terms we get

\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}

\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2

\frac{M_x}{M_y}=(\frac{140}{581})^2

\frac{M_x}{M_y}=0.05806\approx 0.0581

8 0
3 years ago
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