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2 years ago

Where on a roller coaster is centripetal forces at work.

1 answer:

5 0

Regardless of where the cars are in the loop, centripetal force is always directed toward the center of rotation. So even if a car you're riding in is at the top of a loop, upside-down, you will feel pressed into your seat.

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An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 2

ozzi

Answer:

The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have

The temperature T = 602.64K and the final pressure P = 110.24MPa

Explanation:

Numbeer of moles of gas = 3.00 mol

initial volume = 230.8×10−6 m3

final volume = 133.4×10−6 m3 .

released energy = 8240 J

Temperature = Constant = T

Pressure = $p_{f}$ =unknown

From the relation the combined ideal gas law, PV = nRT

Where R = 8.314 4621.JK−1mol−1

we have The release energy from compression P1V1 -P2V2

-qrev = -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 8240J

n = 3

Hence -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 3×8.314 462×ln $(\frac{133.4}{230.8})$ × T= -8240 J

or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K

The Final pressure is given by

PV = n×R×T from where we have V = final volume thus

P = (n×R×T)/V = (3×8.134×602.64)÷(133.4× $10^{-6}$ ) = 110237041.1 N/ $m^{2}$ = 110.237MPa

8 0

3 years ago

4.A 200 kg cannon fires a 5 kg chicken into the air at a velocity of 100

exis [7]

Answer:

-2.5

Explanation:

5kg in grams = 5000

200kg in grams = 200,000

V2 = - (M1/M2) V1

Vc = - (Mh/Mc) Vh

Vc = - (5000/200000) 100

Vc = -2.5 units/second

4 0

2 years ago

As 390 g of hot milk cools in a mug, it transfers 30,000 J of heat to the environment. What is the temperature change of the mil

svetoff [14.1K]

We will apply:
Q = mcΔT; where Q is the heat, m is the mass, c is the specific heat capacity and ΔT is the temperature change.

30,000 = 390 x 3.9 x ΔT
ΔT = 19.7 °C

The temperature change is of 19.7 °C.

7 0

4 years ago

What is the magnitude of the net force needed to bring a 2110 kg car to rest from 18.1 m/s in 7.0 seconds?

lozanna [386]

Here is what we know:
a = ?, s = ?, u = 18.1m/s, v = zero/rest, t = 7.0s, m = 2110kg

(a = acceleration, s= displacement, u = initial velocity, v = final velocity, t = time and m is mass)

Now we choose a kinematic formula. Since we know v, u and t, we will use the formula: v = u+at and rearrange it so that we can find a.
a = v-u/t
a = 0-18.1/7.0 = -2.5857...
therefore, a = -2.6m/s

We have our acceleration, now let’s find the net force. To find the force we use one of Newtons laws of motion.
We will use Newtons second law since it describes what happens when one or more forces act upon an object.
F = ma
F = (2110kg)(-2.6m/s)
F = -5486 kg
Therefore the net force F = -5486 N

4 0

4 years ago

A 10-cm-long thin glass rod uniformly charged to 14.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 14.0 nC are pl

Sonbull [250]

Answer:

A. At a point 1cm from the glass rod, E = 1.36 * 10⁶N/C

B. At a point 2cm from the glass rod, E = 5.17 * 10⁵N/C

C. At a point 3cm from the glass rod, E = 6.993 * 10⁵N/C

Explanation:

Parameters given:

Charge of glass rod, Q = 14nC = 14 * 10⁻⁹ C

Charge of plastic rod, q = 14nC = 14 * 10⁻⁹ C

Distance between both rods = 4.5cm = 0.045

A. Electric field strength at a point 1.0cm (0.01m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

Where

E₁ = electric field strength due to glass rod

E₂ = electric field strength due to plastic rod

E₁ = kQ/0.01²

E₂ = kq/(0.045 - 0.01)² = kq/(0.035)²

E = kQ/0.01² + kq/(0.035)² = k(Q/0001 + q/0.001225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.001) + (14 * 10⁻⁹)/0.001225]

E = 9 * 10⁹[(14 * 10⁻⁵) + (1.143 * 10⁻⁵)]

E = 9 * 10⁹ * 15.143* 10⁻⁵

E = 1.36 * 10⁶N/C

B. Electric field strength at a point 2.0cm (0.02m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.02²

E₂ = kq/(0.045 - 0.02)² = kq/(0.025)²

E = kQ/0.02² + kq/(0.025)² = k(Q/0.0004 + q/0.000625)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.0004) + (14 * 10⁻⁹)/0.000625]

E = 9 * 10⁹[(3.5 * 10⁻⁵) + (2.24 * 10⁻⁵)]

E = 9 * 10⁹ * 5.74 * 10⁻⁵

E = 5.17 * 10⁵N/C

C. Electric field strength at a point 3.0cm (0.03m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.03²

E₂ = kq/(0.045 - 0.03)² = kq/(0.015)²

E = kQ/0.03² + kq/(0.015)² = k(Q/0009 + q/0.000225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.009) + (14 * 10⁻⁹)/0.000225]

E = 9 * 10⁹[( 1.55 * 10⁻⁵) + (6.22 * 10⁻⁵)]

E = 9 * 10⁹ * 7.77 * 10⁻⁵

E = 6.993 * 10⁵N/C

4 0

4 years ago