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Nitella [24]
3 years ago
12

Suppose a 49-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 585 N sits on t

he sled, what force is needed to pull the sled across the snow at constant speed?
Physics
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction,\mu_k=0.11

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

F=\mu_k N

Where N= Normal=mg

F=0.11\times 634=69.74 N

Hence, the force is needed  to pull the sled across the snow at constant speed=69.74 N

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.
Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

  1. Under the assumption that the can is a closed system, the conservation law applied to the system would be: E_{in}-E_{out}=E_{change}, where E_{in} is all energy entering the system, E_{out} is the total energy leaving the system and, E_{change} is the change of energy of the system.
  2. As the purpose is to kept the beverage can at constant temperature, the change of energy (E_{change}) would be 0.
  3. The energy  that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4) where \varepsilon is the emissivity of the surface, \sigma=5.67*10^{-8}\frac{W}{m^2K} known as the Stefan–Boltzmann constant, A_S is the total area of the exposed surface, T_S is the temperature of the surface in Kelvin, T_{\infty} is the environment temperature in Kelvin.
  4. For the can the surface area would be ta sum of the top and the sides. The area of the top would be A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2, the area of the sides would be A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2. Then the total area would be A_{total}=A_{top}+A_{sides}=0.01624m^2
  5. Then the radiation heat transferred to the can would be Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W.
  6. The can would lost heat evaporating water, in this case would be Q_{out}=\frac{dm}{dt}*h_{fg}, where \frac{dm}{dt} is the rate of mass of water evaporated and, h_{fg} is the heat of vaporization of the water (2257\frac{J}{g}).
  7. Then in the conservation balance: Q_{in}-Q_{out}=Q_{change}, it would be1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0.
  8. Recall that 1W=1\frac{J}{s}, then solving for \frac{dm}{dt}:\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}
5 0
2 years ago
If heat is gained from a source, it is __________ from the source.
Strike441 [17]

Answer:

Derived, or expended

4 0
2 years ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
Crank

Answer: 1000 Hz

Explanation:

You can calculate frequency by dividing velocity by wavelength

Frequency = velocity/wavelength

Find velocity first.

900 m/3 s = 300 m/s

Plug values in to find frequency.

F = (300 m/s)/0.3 m

F = 1000 Hz

8 0
3 years ago
Why do the golf ball have dimples?
Igoryamba

Answer:

HELLO! I WILL EXPLAIN WHY GOLF BALLS HAVE DIMPLES.

Explanation:

DIMPLES ON A GOLF BALL CREATE A THIN BOUNDARY LAYER OF AIR THAT CLINGS TO THE BALL'S SURFACE. THIS ALLOWS THE AIR TO FOLLOW THE BALL'S SURFACE A LITTLE FARTHER AROUND THE BACKSIDE OF THE BALL, THEREBY DECREASING THE SIZE OF THE WAKE.

I AM SO SORRY ABOUT THE CAP LOCKS I THINK MY CAP LOCKS BUTTON IS PUSHED, AND IT IS NOT POPPING UP FOR SOME REASON.

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2 years ago
Read 2 more answers
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

3 0
3 years ago
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