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docker41 [41]
3 years ago
5

gives the speed v versus time t for a 0.500 kg object of radius 6.00 cm that rolls smoothly down a 30° ramp. The scale on the ve

locity axis is set by vs = 4.0 m/s. What is the rotational inertia of the object?

Physics
1 answer:
ddd [48]3 years ago
8 0

Answer:

Rotational inertia of the object is given as

I = 7.2 \times 10^{-4} kg m^2

Explanation:

As we know that the acceleration of the object on inclined plane is given as

a = \frac{gsin\theta}{1 + k^2/R^2}

now we know that velocity at any instant of time is given as

v = at

now we know that if the graph between velocity and time is given then the slope of the graph will be same as acceleration

so here we have

\frac{gsin\theta}{1 + k^2/R^2} = slope

now from the graph slope of the graph is given as

slope = \frac{3.5 - 0}{1}

\frac{gsin\theta}{1 + k^2/R^2} = 3.5

\frac{9.81 sin30}{1 + k^2/R^2} = 3.5

k^2 = 0.4 R^2

now rotational inertia is given as

I = mk^2

I = 0.5(0.4)(0.06)^2

I = 7.2 \times 10^{-4} kg m^2

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 61 m/s^2. If he reaches the gr
ira [324]

Answer:

55.66 m

Explanation:

While falling by 50 m , initial velocity u = 0

final velocity = v , height h = 50 , acceleration g = 9.8

v² = u² + 2gh

= 0 + 2 x 9.8 x 50

v = 31.3 m /s

After that deceleration comes into effect

In this case final velocity v = 17 m/s

initial velocity u = 31.3 m/s

acceleration a = - 61 m/s²

distance traveled h = ?

v² = u² + 2gh

(17)² = (31.3)² - 2x 61xh

h = 690.69 / 2 x 61

= 5.66 m

Total height during which he was in air

= 50 + 5.66

= 55.66 m

3 0
2 years ago
the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
3 years ago
A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a
trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden d_1=1.1cm

Speed of water in the hose is v_1=0.95m/sec

Number of holes n = 22

Diameter of each holes d_2=15cm

According to continuity equation A_1v_1=A_2v_2

d_1^2\times v_1=22\times d_2^2v_2

1.1^2\times 0.95=22\times 0.15^2\times v_2

v_2=2.322m/sec

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0
2 years ago
A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this
baherus [9]
20 characters longer later and woah
5 0
3 years ago
In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at
Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0
3 years ago
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