Answer: the speed at which it falls toward the Earth.
Explanation:
A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.
Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).
This kind of motion is the result of two independent motions: vertical motion and horizontal motion.
The observed net velocity is the vectorial sum of the vertical and horizontal velocities.
The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).
The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.
Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.
An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:
1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage. This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.
2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.
3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP. This newly made ATP can now be broken down to issue energy to fuel activity.
Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As
As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So
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
A.
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
B.
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.
Answer:
halve the slit separation
Explanation:
As we know that
In YDS experiment, the equation of fringe width is as follows
where,
D denotes the separation in the middle of screen and slits
d denotes the distance in the middle of two slits
And to increase the Δx we have to decrease the d i.e, the distance between the two slits
Hence, the first option is correct
