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Tamiku [17]
1 year ago
11

A uniform crate with a massof 30 kg must be moved up along the 15° incline without tipping. Knowing that force P is horizontal,

determine (a) the largest allowable coefficient of static friction between the crate and the incline, (b) the corresponding magnitude of force P.
Physics
1 answer:
lakkis [162]1 year ago
8 0

Answer:

coefficient of friction =0.268

magnitude of force P=289.78N

Explanation:

The coefficient of friction is obtained by mgsinФ/mgcosФ=tanФ=tan15=0.268

force P is horizontal as stated in the question, horizontal component of P=mgcosФ=30*10*cos15=289.78

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A skateboarder rides swiftly up the edge of a bowl-shaped surface and leaps into the air. While in the air, the skateboarder fli
Katen [24]

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of kgm^{2 }\  / s.

  • In the given question the skateboarder rides quickly up the bottom of a bowl-shaped surface and climb into the air.it means it is rotational movement also it is not touching anything so it is angular momentum.
  • All the other option is incorrect because it is not follows the given scenario
3 0
1 year ago
3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag
aleksandr82 [10.1K]

Answer:

(a) W=-19.25J

(b) W=-52.8J

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

W=nRTln(\frac{V_1}{V_2} )

Whereas the moles are computed from the given mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Thus, the work is:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Regards.

4 0
1 year ago
The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a
Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

J=F.\Delta t

J=10^6\ N\times 20\ s

J=2\times 10^7\ N-s

(b) Impulse is also given by the change in momentum i.e.

J=\Delta p=p_f-p_i

J=p_f-p_i

p_f=J+p_i

p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s

p_f=20143000\ kg-m/s

(c) For new velocity,

v_f=\dfrac{p_f}{m}

v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}

v_f=1549.46\ m/s

Hence, this is the required solution.

3 0
1 year ago
Read 2 more answers
Please help meee<3~I have no idea what the answer is~
valina [46]
9).
In a properly conducted experiment, the experimenter controls one part
of the experiment to see what the other parts do.

Example:  Experiment to describe the effect of heat on ice.
Take two same-size ice cubes out of the same ice tray in the same fridge.
Place each one on a little temperature-controlled electric pad.
Turn one pad on, to make it warm.  Leave the other pad turned off. 
You CONTROL one part of the experiment:  the amount of heat that
       the ice cube gets.
You KNOW that the heat is the only thing different between the two
     ice cubes.  They're the same size.  They were both made from
     the same water, and froze in the same tray in the same fridge.
so
You KNOW that any difference will be the result of the heat on one of them.
You WATCH to see what happens to the one that gets the heat.


10).
An hypothesis is a prediction of what you believe may be true. 
Once you have it, it's time to do an experiment to find out whether
your hypothesis is true.

Example:
I have an hypothesis.  It predicts that when ice gets warm it melts.
Experiment:
Take two same-size ice cubes out of the same ice tray in the same fridge.
Set one ice cube down on the table.
Keep the other one in your hand.
The one in your hand melts while the one on the table is still solid.
Is the hypothesis correct ?
Maybe it is.  Maybe it isn't.
We know that there's something about your hand that melts ice.
It may be the warmth.  But it may be something else about human skin.
You'll need another experiment, slightly different, to find out if it's the warmth.

4 0
1 year ago
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

6 0
1 year ago
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