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2 years ago

A 20.0 kg curling stone travels 30.0 m along the ice surface. If the frictional force is 10.0 N, the thermal energy produced is

Physics

2 answers:

raketka [301]2 years ago

5 0

Answer:

thermal energy is 300 J

Explanation:

Thermal energy produced while moving on the rough floor is same as the work done by friction force

because friction force is a non conservative force which will produce thermal energy in the form of energy loss

so here this thermal energy is given by

$E = W_f$

$E = F_f . d$

here we have

$F_f = 10 N$

$d = 30 m$

so the work done is given by

$E = 10(30) = 300 J$

So the thermal energy is 300 J

evablogger [386]2 years ago

4 0

The thermal energy is where the work of friction comes from. That is what stops it eventually. In this case a counter force of 10N is applied over the distance of 30.0m. The energy is given by Force*Distance. Here this is 300J. This friction work is the thermal energy.

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A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

erastovalidia [21]

Would presume you are asked to find the volume, since there is no second volume.

By General Gas Law:

P₁V₁/T₁ = P₂V₂/T₂

1.6 * 168 /255 = 1.3*V₂/285

V₂ = 1.6 * 168 * 285 / (1.3*255)

V₂ = 231.095

Final volume ≈ 231 cm³

7 0

2 years ago

Read 2 more answers

You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Fris

Usimov [2.4K]

Answer:

Torque τ =w ×0 = 0

Explanation:

We know that the torque is given by the product of the force and perpendicular distance between the force and the axis.

Here the gravity force act at the center and the rotational axis is also passing through the center.

Therefore the perpendicular distance between the force and the rotational axis would be zero.

Hence the torque will be

Torque = Force × Perpendicular distance

Torque = mg×0 = 0

Therefore the torque would be zero.

7 0

2 years ago

A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with

Lana71 [14]

Answer:

h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

Em₀ = K = ½ m v₁²

final point. Highest point of the curl

$Em_{f}$ = U = m g y

Find the height y = 2R

Em₀ = Em_{f}

½ m v₁² = m g 2R

v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

-fr = m a (1)

Y Axis

N - W = 0

N = mg

the friction force has the formula

fr = μ N

fr = μ m g

we substitute 1

- μ mg = m a

a = - μ g

having the acceleration, we can use the kinematic relations

v² = v₀² - 2 a x

v₀² = v² + 2 a x

the length of this zone is x = 2R

let's calculate

v₀ = √ (4 gR + 2 μ g 2R)

v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

Em₀ = U = m g h

final point. Just before reaching the area with rubbing

$Em_{f}$ = K = ½ m v₀²

Em₀ = Em_{f}

mgh = ½ m 4gR(1 + μ)

h = ½ 4R (1+ μ)

h = 2 R (1 +μ)

7 0

2 years ago

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu

Dmitrij [34]

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system= $4.5\times 10^5$ J

Heat transfer into the system= $U_1=3.2\times 10^6$ J

Heat transfer to the environment= $U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy = $-3.25\times 10^6 J$

7 0

2 years ago

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob

givi [52]

Answer:

Charge, $Q=1.56\times 10^{-8}\ C$

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

$E=\dfrac{kQ}{r^2}$

Q is the charge on an object

$Q=\dfrac{Er^2}{k}$

$Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}$

$Q=1.56\times 10^{-8}\ C$

So, the charge on the object is $1.56\times 10^{-8}\ C$ . Hence, this is the required solution.

7 0

2 years ago