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3 years ago

First, record some data for this comparison in the table below.

Physics

1 answer:

kondor19780726 [428]3 years ago

5 0

Answer:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).

Explanation:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

This is the answer

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A 16 g rifle bullet traveling 240 m/s buries itself in a 3.6 kg pendulum hanging on a 2.5 m long string, which makes the pendulu

iren [92.7K]

Answer:

x = 0.54 m

y = 0.058 m

Explanation:

m = mass of the bullet = 16 g = 0.016 kg

v = speed of bullet before collision = 240 m/s

M = mass of the pendulum = 3.6 kg

L = length of the string = 2.5 m

h = height gained by the pendulum after collision

V = speed of the bullet and pendulum combination

Using conservation of momentum

m v = (m + M) V

(0.016) (240) = (0.016 + 3.6) V

V = 1.062 m/s

Using conservation of energy

Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination

(m + M) g h = (0.5) (m + M) V²

(9.8) h = (0.5) (1.062)²

h = 0.058 m

y = vertical displacement = h = 0.058 m

x = horizontal displacement

horizontal displacement is given as

x = sqrt(L² - (L - h)²)

x = sqrt(2.5² - (2.5 - 0.058)²)

x = 0.54 m

8 0

3 years ago

A uniform 2.00-kg circular disk of radius 20.0 cm is rotating clockwise about an axis through its center with an angular speed 3

riadik2000 [5.3K]

Answer:

w = 132.57 rad / s

Explanation:

To solve this exercise we must define a system formed by the two discs, therefore the angular momentum is conserved

initial

L₀ = I₀ w₀

final

L_f = I w

how the moment is preserved

L₀ = L_f

I₀ w₀ = I w

the moment of inertia of disk 1 is

I₀ = ½ m₁ r₁²

The moment of inertia of the set is

I = I₀ + I₂

I = ½ m₁ r₁² + ½ m₂ r₂²

we substitute

½ m₁ r₁² w₀ = (½ m₁ r₁² + ½ m₂ r₂²) w

w = $\frac{ m_1r_1^2 }{ m_1 r_1^2+ m_2r_2^2} \omega_o$

let's reduce the magnitudes to the SI system

w₀ = 30.0 rev / s (2π rad / 1 rev) = 188.5 rad / s

let's calculate

w = ( $\frac{ 2 \ 0.20^2}{ 2 \ 0.20^2 + 1.5 \ 0.15^2 }$ ) 188.5

w = ( $\frac{0.08}{ 0.11375}$ ) 188.5

w = 132.57 rad / s

4 0

3 years ago

3. A car is traveling up a 3% grade, with the speed of 85mph, on a road that has good, wet pavement. A deer jumps out onto the r

solniwko [45]

Answer:

Explanation:

From the given information;

Let assume that:

the wheel radius = 15 inches

the driveline slippage = 3%; &

the gear reduction ratio (overall) = 2.5 to 1

So; using the equation:

$v_1= \dfrac{2 \pi r n_o (1 -i)}{\varepsilon_o}$

$v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (1 -0.03)}{2.5}$

$v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (0.97)}{2.5}$

$v_1 = 126.92 \ fp^3$

$frl = 0.01 ( 1+ \dfrac{v}{147}) \ if \ v \ is \ ft/sec$

$frl = 0.01 \Bigg( 1+ \dfrac{\dfrac{126.92 +(20)1.47 }{2} }{147}\Bigg)$

$S = \dfrac{v_b ( v_1^r-v_2^r)}{2g(n_b \mu + frl \pm sin \ y}$

where;

$\mu = 0.6$

$291 = \dfrac{1.64( 126.92^2-29.9^2)}{64.4(n_b \times 0.6 +0.01532 +0.03}$

$n_b = 1.33 \to which \ is \ not \ possible$

However;

$n_b \mu = 1.33(0.6) = 0.80$

$\mu = 0.9 \to$ if the car's anti-clock breaking system did not fail

Thus;

$n_b (0.9) = 0.80$

$n_b =\dfrac{ 0.80}{(0.9) }$

$n_b = 0.89$

Hence, the distance is possible if the anti-clock breaking system did not fail.

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3 years ago

What can scientist learn from studying sedimentary rocks?

cricket20 [7]

They are made of rocks

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3 years ago

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vesna_86 [32]

Answer:

The airfoil shape and wing size will both affect the amount of lift. The ratio of the wing span to the wing area also affects the amount of lift generated by a wing. ... The lift then depends on the velocity of the air and how the object is inclined to the flow. Air: Lift depends on the mass of the flow.

Please give brainliest!!!

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3 years ago