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3 years ago

At what vertical velocity should no object be launched at in order to achieve a height of 20m?

Physics

1 answer:

Free_Kalibri [48]3 years ago

7 0

Answer:

19.8 m/s

Explanation:

Given:

Maximum vertical displacement of the object (H) = 20 m

Acceleration due to gravity (g) = 9.8 m/s²

At maximum height, the velocity of the object is 0 m/s for a moment. So, final velocity (v) at the maximum height is 0 m/s.

Now, let the initial velocity or velocity at launch be 'u' m/s.

Now, using the following equation of motion for vertical motion:

$v^2=u^2-2gH$

Rewriting in terms of 'u', we get:

$u^2=v^2+2gH\\\\u=\sqrt{v^2+2gH}$

Plug in the given values and solve for 'u'. This gives,

$u=\sqrt{0+2\times 9.8\times 20}\\\\u=\sqrt{392}\\\\u=19.8\ m/s$

Therefore, the vertical velocity at the launch is 19.8 m/s.

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What is the mass of an object that has a weight of 80.0 N?

Ira Lisetskai [31]

Explanation:

SUPONIENDO QUE LA ACELERACIÓN DE LA GRAVEDAD ES $9.80 m/s^{2}$

USANDO LA SEGUNDA LEY DE NEWTON:

<em>m</em> = 80.0 N/ $9.80 m/s^{2}$ = 8.16 kg

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3 years ago

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Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi

PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

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2 years ago

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{c}{f}$

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

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2 years ago

Can someone please help me?

svetlana [45]

Answer:

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3 years ago

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Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1

miv72 [106K]

Answer:

w = 4,786 rad / s , f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

L₀ = 0 + I₂ w₂

Final after coupling

$L_{f}$ = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

I = ½ M R²

How the moment is preserved

L₀ = $L_{f}$

I₂ w₂ = (I₁ + I₂) w

w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

d₁ = 60 cm = 0.60 m

d₂ = 40 cm = 0.40 m

f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

w₂ = 2 π f₂

w₂ = 2π 3.33

w₂ = 20.94 rad / s

Let's replace

w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

w = 20.94 1.28 / 5.6

w = 4,786 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 4.786 / 2π

f = 0.76176 Hz

7 0

3 years ago

At what vertical velocity should no object be launched at in order to achieve a height of 20m?​

At what vertical velocity should no object be launched at in order to achieve a height of 20m?