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3 years ago

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34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf

orm is 5 m away from the door when the train starts to pull away and heads toward the door at an acceleration of 1.2 m/s2. How long does it take the passenger to reach the door?

1 answer:

ira [324]3 years ago

3 0

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

$d_1 = \frac{1}{2}(0.6) t^2$

also the distance moved by passenger

$d_2 = \frac{1}{2}(1.2) t^2$

so we will have

$d_1 + 5 = d_2$

$0.3 t^2 + 5 = 0.6 t^2$

$0.3 t^2 = 5$

t = 4.08 s

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A car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s. The car continues for 5.00 s at con

barxatty [35]

Answer:

19.5 m

Explanation:

Using the equation of motion,

For the first 3 seconds,

s = ut +1/2at².......................... Equation 1

Where t = time, u = initial velocity, a = acceleration, s = distance

Note: Since the car starts from rest, u = 0 m/s.

Given: a = 1.0 m/s, t = 3 s, u = 0 m/s

Substitute into equation 1

s = 0×3+1/2(3)²(1)

s = 9/2

s = 4.5

In the remaining five seconds, at a constant velocity,

s' = vt............. Equation 2

Where v = velocity, t = time.

Recall,

v = u+at

v = 0+1(3)

v = 3 m/s.

Also, t = 5 s.

Substitute into equation 2

s' = 3(5)

s' = 15 m.

Total distance = 15+4.5

Total distance = 19.5 m.

Hence the car travels 19.5 m

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3 years ago

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque

stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

$\lambda _1=589 nm$

$\lambda _2=589.6 nm$

Energy is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

x=velocity of Light

$\lambda$ =wavelength of emission

$E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}$

$E_1=3.374\times 10^{-19} J$

$E_1 in kJ/mol$

$E_1=203.2 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}$

$\nu _1=5.09\times 10^{14} Hz$

Energy corresponding to $\lambda _2$

$E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}$

$E_2=3.371\times 10^{-19} J$

$E_2=203.02 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}$

$\nu _1=5.088\times 10^{14} Hz$

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3 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Nadya [2.5K]

Answer:

1822.14 J/kg C

Explanation:

Mass of aluminum = 0.100 kg

Mass of Water = 0.250 kg

Mass of copper = 0.300 kg

Mass of unknown object = 0.70 kg

Initial temperature of Aluminum = 10 C

Initial temperature of water = 10 C

Initial temperature of copper = 30 C

Initial temperature of = 100 C

Final temperature of all substances = 20 C

Change in temperature of each of them Aluminum, water , copper and unknown material respectively = 10 C , 10 C , -10 C, -80 C respectively

Specific heat of each of them Aluminum, water , copper and unknown material respectively = 900,4186,387 and c respectively.

Heat gained by aluminum and water = (0.1)(900)(10) + (0.250)(4186)(10) = 11365 j

heat lost by copper and unknown material =(0.3)(387)(-10) = -1161

heat lost by unknown material = (0.070)(c)(-80) = -5.6 c

Now solve for c using the calorimetry principle.

Heat lost + Heat gained = 0

⇒ -5.6 c = -11365 - (-1161) = -10204 j

Specific heat of unknown material = - 10204 / -5.6

= 1822.14 j/kg C

b) Material having a specific heat of 1800 j /kg C is

Polyurethane elastomer.

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3 years ago

All endothermic reactions absorb heat and make the surroundings warm

svet-max [94.6K]

Answer:

b: false

Explanation:

It's true that endothermic reactions absorb heat from their surrounding. But this makes the surroundings cooler, not warmer.

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3 years ago

Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

Step2247 [10]

Answer:

$\Delta p = 90.7 kPa$

Explanation:

specific gravity of oil is $= \frac{\rho_{oil}}{\rho_w}$

$\rho_{oil} = 0.85*1000 = 850 kg/m3$

we know that

change in pressure for oil is given as

$\Delta p = \rho gh$

here density and h is for oil

$\Delta p = 850*5 *9.81 = 41,692.5 kPa$

change in pressure for WATER is given as

$\Delta p = \rho gh$

here density is for water and h is for water

$\Delta p = 1000*5 *9.81 = 49,050 kPa$

pressure change due to both is given as

$\Delta p = 41692.3 + 49050 = 90742.5 N/m2$

$\Delta p = 90.7 kPa$

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3 years ago

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