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Delicious77 [7]

3 years ago

14

A roller coaster car is elevated to a height of 30 m and released from rest to roll along a track. At a certain time T it is at

a height of 2 m and has lost 25.000 J of energy to friction. The car has a mass of 800 kg. Answer the following questions. (a) How fast is the car going at time T? (b) How fast would the car be going at time T if the track were frictionless?

1 answer:

Naddika [18.5K]3 years ago

4 0

Explanation:

Initial energy = final energy + work done by friction

PE = PE + KE + W

mgH = mgh + 1/2 mv² + W

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000

v = 22.1 m/s

Without friction:

PE = PE + KE

mgH = mgh + 1/2 mv²

(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²

v = 23.4 m/s

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A thermometer should be air dried so as to not damage it or anything similar. The three steps to cleaning it are washing, rinsing, and air drying it after. You shouldn't try to dry it other ways because it can damage it and this can cause a lot of troubles, so things cold air blowers or similar things can be very good in cleaning your thermometer.

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3 years ago

One mole of titanium contains how many atoms?

Vladimir [108]

Answer:

$\boxed {\boxed {\sf D. \ 6.02*10^{23} \ atoms}}$

Explanation:

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Regardless of the particles, there will always be <u>6.02*10²³</u> (also known as Avogadro's Number) particles in one mole of a substance.

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3 0

3 years ago

Read 2 more answers

A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is

Tpy6a [65]

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

Force applied = 100 N
Distance covered = 5 metres
Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

$\bf \boxed{Work = Force \times displacement}$

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

$\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}$

Substituting values in above formula;

Power = 500/ 10

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Hence, power = 50 W .

5 0

3 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

2 years ago

How much energy is required to raise the temperature of 50.0 grams of water 10.0 degree C? (Explain yourself answer in joules!)

maria [59]

The amount of energy needed is 2093 J

Explanation:

The amount of energy needed to increase the temperature of a substance by $\Delta T$ is given by the equation

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is the increase in temperature

For the water in this problem, we have

m = 50.0 g = 0.050 kg

$C=4186 J/g^{\circ}C$ (specific heat capacity of water)

$\Delta T=10.0^{\circ}C$

Therefore, the amount of energy needed is

$Q=(0.050)(4186)(10)=2093 J$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

4 0

2 years ago