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MissTica
2 years ago
9

What is always included in a comparative investigation?

Physics
2 answers:
12345 [234]2 years ago
5 0

Answer:

A comparison is always included in a comparative investigation. Comparative investigation collects data about different objects, features, organisms, sometimes under a different set of conditions to compare them. The similarities and differences that occur during a set of time or conditions are observed and compared.

Explanation:

iragen [17]2 years ago
4 0
A comparison is always included in a comparative investigation. Comparative investigation collects data about different objects, features, organisms, sometimes under a different set of conditions to compare them. The similarities and differences that occur during a set of time or conditions are observed and compared.
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Air at 25°C and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligible
Vsevolod [243]

Answer:

T_{2} = 25^{\circ}C

Solution:

As per the question:

Temperature, T_{1} = 25^{\circ}C

Pressure, P_{1} = 5 atm[\tex]Now,We know that the process of throttling occurs in adiabatic conditions where no heat is transferred in between the system and the surrounding, i.e., Q = 0Thus no work is done in this process, i.e., W = 0Enthalpy also remains same from one state to the other state, i.e., [tex]\Delta h = 0

Therefore, from the eqn:

Q - W = \Delta h + \Delta KE

We can write:

\Delta h = h_{1} - h_{2} = 0

\Delta h = C_{p}\Delta T

C_{p}(T_{2} - T_{1}) = 0

Thus

T_{1} = T_{2} = 25^{\circ}C

where

T_{2} = Exit temperature

4 0
2 years ago
A transverse wave is set up in a very long string. The oscillator is set at 20.0 Hz, and the wave speed is 78 m/s. The amplitude
Sedbober [7]

To solve this problem we must apply the concepts related to Tangential Acceleration based on angular velocity and acceleration, and therefore, we must also calculate angular velocity based on the given frequency. For all these problems we will take the Units to the International System. The maximum acceleration would then be defined as,

a_{max} = \omega^2 A

Here,

\omega= Angular velocity

A = Amplitude

At the same time the angular velocity is described as,

\omega = 2\pi f

Here f means the frequency of the wave. Substituting,

\omega = 2 \pi (20)

\omega = 40\pi

A = 5.2cm

A = 0.052m

Replacing at the first equation,

a_{max} = (40\pi )^2 (0.052)

a_{max} = 821.15m/s^2

Therefore the maximum particle acceleration for a point on the string is 821.15m/s^2

6 0
3 years ago
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert
WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h

\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h

h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g} (1)

Where:

h - Maximum height of the wood block, in meters.

v - Initial speed of the block, in meters per second.

\mu_{k} - Kinetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

m - Mass, in kilograms.

s - Distance travelled by the wood block along the wooden ramp, in meters.

\theta - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that v = 10\,\frac{m}{s}, \mu_{k} = 0.20 and g = 9.807\,\frac{m}{s^{2}}, then the height reached by the block above its starting point is:

h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

h = 4.249\,m

The wood block reaches a height of 4.249 meters above its starting point.

5 0
2 years ago
Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is
n200080 [17]

Answer:

90 degree

Explanation:

Electric field line is vertical to the electric field line.

4 0
2 years ago
Un automovil transita por una curva en forma de U y recorre una distancia de 400m en 30s sin embargo su posición final está a so
pochemuha

Answer:

Definimos:

Rapidez media es igual al cociente entre la distancia recorrida y el tiempo que se tarda en recorrer esa distancia.

En este caso la distancia recorrida es 400m, y el tiempo que se tarda es 30s, entonces la rapidez media va a ser:

RM = 400m/30s = 13.33 m/s

La velocidad media por otro lado, es igual al cociente entre el desplazamiento y el tiempo necesario para desplazarse.

El desplazamiento es igual a la distancia entre la posición final y la posición inicial, que en este caso eso 40m, y el tiempo necesario sigue siendo 30s, entonces la velocidad media va a ser:

VM = 40m/30s = 1.33 m/s

8 0
3 years ago
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