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2 years ago

How much heat is absorbed by 34g iron skillet its temperature rises from 12c to 24 c ?

Physics

1 answer:

Mrrafil [7]2 years ago

7 0

Answer:

$1.84\cdot 10^5 J$

Explanation:

The heat absorbed by the piece of iron is given by:

$Q=mC_S \Delta T$

where:

m = 34 g is the mass of the piece of iron

$C_s = 450 J/g^{\circ}C$ is the specific heat of iron

$\Delta T=24 C-12 C=12^{\circ}C$ is the variation of temperature of the piece of iron

Substituting numbers into the formula, we find

$Q=(34 g)(450 J/gC)(12 C)=1.84\cdot 10^5 J$

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You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle

AlekseyPX

Answer:

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

$- mgh - F_f (s) = 0 - \frac{1}{2}mv^2$

here we know that

$F_f = \mu_k mg cos\theta$

also we know that the length of the incline is given as

$s = \frac{h}{sin\theta}$

now we have

$- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2$

so we have

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

3 0

2 years ago

What do you mean by kinetic energy and potential energy ?

Verizon [17]

Kinetic Energy of an object is the measure of the work an object can do by virtue of its motions..

Formula - KE = 1/2 mv^2

Where KE is the kinetic energy, m is the body’s mass, and v is the body’s velocity.

Potential energy is the stored energy in any object or system by virtue of its position or arrangement of parts..

Formula - W = mgh

Where,

. m is the mass in kilograms

. g is the acceleration due to gravity

. h is the height in meters

Hope it helpz~ uh..

4 0

1 year ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

3 years ago

What do stars begin as

Valentin [98]

Cloud of dust and gas!

7 0

3 years ago

Read 2 more answers

A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st

Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

$\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r$

where,

$\tau$ = shear stress at a distance 'r' from the center

T = is the applied torque

$I_{p}$ = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

$\tau _{max}=\frac{4}{3}\times \frac{V}{A}$

Applying values we get

$\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa$

3 0

2 years ago