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2 years ago

A stock person at the local grocery store has a job consisting of the following five segments:

Physics

1 answer:

vaieri [72.5K]2 years ago

6 0

Answer:

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

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A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause

ANEK [815]

Answer:

The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid

B. The contact area of each tire with the track.

C. The co-efficent of static friction between the tires and the track.

D. The co-efficent of static friction between the tires and the track.

Explanation:

4 0

1 year ago

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0

2 years ago

Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to

marta [7]

Answer:

$\frac{ 112.5}{15+m_{A}}=v_{f}$

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed $v_{iA}$ of 0 m/s and a mass $m_{A}$ and the IMAX camera with an initial speed $v_{ic}$ of 7.5 m/s and a mass $m_{c}$ of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

$P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}$

By the law of conservation we know that $P_{i} =P_{f}$

For $P_{f}$ (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

$P_{i} =P_{f}=112.5\\$

$112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}$

3 0

1 year ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$