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3 years ago

The chemical symbol H represents which of the following elements?

Physics

2 answers:

olga55 [171]3 years ago

7 0

Answer:
Hydrogen

Explanation:

Helen [10]3 years ago

6 0

Answer:

Hydrogen

Explanation:

Hydrogen is a chemical element with chemical symbol H and atomic number 1. It also has an atomic weight of 1.008 u

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A typical reaction time to get your foot on the brake in your car is 0.2 second. If you are traveling at a speed of 60 mph (88 f

katrin [286]

Explanation:

Not enough information. It really depends on the technical details of the car ( the data provided is offering just the human factor of the reaction, not the time for getting the impulse through when using the breaks

5 0

3 years ago

Read 2 more answers

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

A trough is 10 feet long and its ends have the shape of isosceles triangles that are 3 feet across at the top and 1 foot high. I

tensa zangetsu [6.8K]

Answer:

1.6 ft/min

Explanation:

Since trough is 10 ft long and water is filled at the rate of 12ft3/min. We can calculate the rate of water filled with respect to area:

= 12 / 10 = 1.2ft2/min

As the water level rises, so does the water surface, or the bottom side of the isosceles triangles. In fact we can calculate the bottom side when the trough is half foot deep:

= 3 / 2 = 1.5 ft

The rate of change in water level would be the same as calculating the height of the isosceles triangles knowing its base

= 1.2 * 2 / 1.5 = 1.6 ft/min

6 0

3 years ago

Two 84.5 g ice cubes are dropped into 30 g of water in a glass. If the water is initially at a temperature of 50 C and if the ic

Setler [38]

Answer:

final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 182 g$

total amount of water

$m_{water} = 17 g$

Explanation:

After thermal equilibrium is achieved we can say that

Heat given by water = heat absorbed by ice cubes

so we will have

Heat given by water to reach 0 degree C

$Q_1 = m_1s_1 \Delta T_1$

$Q_1 = 0.030(4186)(50 - 0)$

$Q_1 = 6279 J$

heat absorbed by ice cubes to reach 0 degree

$Q_2 = m_2 s_2 \Delta T_2$

$Q_2 = (0.169)(2100)(30)$

$Q_2 = 10647 J$

so we will have

$Q_2 > Q_1$

so here we can say that few amount of water will freeze here to balance the heat

$10647 - 6279 = mL$

$m = \frac{10647 - 6279}{335000}$

$m = 13 g$

so final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 84.5 + 84.5 + 13$

$m_{ice} = 182 g$

total amount of water

$m_{water} = 30 - 13$

$m_{water} = 17 g$

4 0

3 years ago

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

Archy [21]

Answer:

a) y= 3.5 10³ m, b) t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

y₁ = y₀ + v₀ t + ½ a t²

y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

y₁ = ½ 16 10²

y₁ = 800 m

At the end of this stage it has a speed

v₁ = vo + a₁ t₁

v₁ = 0 + 16 10

v₁ = 160 m / s

Stage 2

y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

y₂ = 800 + 150 5 + ½ 11 5²

y₂ = 1092.5 m

Speed is

v₂ = v₁ + a₂ t

v₂ = 160 + 11 5

v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

v₃² = v₂² - 2 g y₃

0 = v₂² - 2g y₃

y₃ = v₂² / 2g

y₃ = 215²/2 9.8

y₃ = 2358.4 m

The total height is

y = y₃ + y₂

y = 2358.4 + 1092.5

y = 3450.9 m

y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

v₃ = v₂ - g t₃

v₃ = 0

t₃ = v₂ / g

t₃ = 215 / 9.8

t₃ = 21.94 s

The time to climb is

$t_{s}$ = t₁ + t₂ + t₃

t_{s} = 10+ 5+ 21.94

t_{s} = 36.94 s

The time to descend from the maximum height is

y = v₀ t - ½ g t²

When it starts to slow down it's zero

y = - ½ g t_{b}²

t_{b} = √-2y / g

t_{b} = √(- 2 (-3450.9) /9.8)

t_{b} = 26.54 s

Flight time is the rise time plus the descent date

t = t_{s} + t_{b}

t = 36.94 + 26.54

t =63.84 s

t = 64 s

3 0

3 years ago