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Mazyrski [523]
4 years ago
11

Analyze 1. What happened to the time it took for the marble to reach the table (t) as the angle of incline increased? 2. What is

the relationship between the acceleration of the marble (a) and the time it takes the marble to reach the table (t)? Draw Conclusions 1. According to your graph, what would be the acceleration of the marble at 90°? 2. Given what you know about the acceleration of Earth's gravity (g = 9.8 m/s2), is this number accurate? If not, explain why you think it is not accurate. Explore Further 1. What factors might affect the accuracy of this experiment? Describe one way you could improve the experiment's design. List any equipment you might use, and explain how you would use it. 2. If the acceleration of gravity is 9.8 m/s2, then that means an object falling at 90° will be traveling 9.8 m/s after 1 second, 19.6 m/s after 2 seconds, and so on. Use the data values in your table to sketch a rough graph of velocity versus time for the 10° angle and another for the 40° angle. What value do the slopes of these graphs represent? Which graph has the greater slope? Why?

Physics
1 answer:
liraira [26]4 years ago
6 0
It traveled and gained kinetic energy
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Its like newtons 3rd law that once in motion a outer force has to stop it
5 0
3 years ago
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A 1502.7 kg car is traveling at 33.1 m/s when
Bond [772]

By definition of average acceleration,

<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.

5 0
3 years ago
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
3 years ago
A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and
Airida [17]

Answer:

Explanation:

Given

N_1=1 rev/s

angular velocity \omega =2\pi N_1=6.284 rad/s

Combined moment of inertia of stool,student and bricks =6\ kg.m^2

Now student pull off his hands so as to increase its speed to suppose N_2 rev/s

\omega _2=2\pi N_2  

After Pulling off hands so final moment of inertia is

I_2=2\ kg-m^2

Conserving angular momentum  as no external torque is applied

I_1\omega _1=I_2\omega _2

6\times 6.284=2\times \omega _2

\omega _2=18.85\ rad/s

N_2=3 rev/s

7 0
3 years ago
Despite a very strong wind, a tennis player
Gnoma [55]

Answer:

Option 5. 1 and 3

Solution:

The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.

The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction  before it hits the ground.

6 0
3 years ago
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