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Scilla [17]
3 years ago
13

A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the

skier encounters a horizontal surface. The skier travels on the horizontal surface a distance of 95.2 meters before coming to a rest. What is the magnitude of the (constant) friction force (in N) that the rough horizontal surface exerts on the skier
Physics
1 answer:
ehidna [41]3 years ago
7 0

Answer: 170.67 N

Explanation:

Given

Mass of skier is m=82.9\ kg

Height of the inclination is h=20\ m

Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.

\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N

Thus, the horizontal friction force is 170.67 N.

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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

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.

5 0
3 years ago
What conversion factor is used to convert 20 cm to m
Anastaziya [24]
The conversion factor you use is 100 cm = 1 m.
You can divide 20 by 100 to get the answer.
20 cm/100 cm =.2 m
Hope this helped!
7 0
3 years ago
20 points and brainliest‼️‼️‼️‼️
Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

3 0
3 years ago
Current evidence suggests that many massive jovian planets orbit at very close orbital distances to their stars. How do we think
ivanzaharov [21]

Answer:

In the Solar system, the Jovian planets are farther from the Sun. Majority of the extrasolar Jovian planets are closer to their stars. These are known as "Hot Jupiters". From the studies, the reason for the existence of massive Jovian planets to be closer to their star is found to be the gravitational interaction of these planets with other massive planets which pushes them closer to their stars. These planets are formed beyond the frost line initially but later on migrate inwards.

4 0
3 years ago
Which of the following physical properties would tell you about an element's mass per unit volume ratio?
Igoryamba
Well I can't see the following physical properties you talked about in the question.

Mass per unit volume ratio is called density.
6 0
3 years ago
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