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3 years ago

The road study guide what does the man look for with his binoculars

1 answer:

3 0

In the story, The Road, the man is looking for food and supplies with his binoculars. There is a boy that is traveling with the man and they are starving. They continue traveling until they see a big house and they enter it to look for food.

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calculate the electric field strength between two parallel plates separated by 0.50 cm, across which is a potential of 12 volts.

Alexxandr [17]

Formula for Electric Field strength , E = V/d

where V = Voltage in volts, and d = distance of separation in meters.

d = 0.5 cm = 0.005 m, V = 12 V

E = V/d = 12 / 0.005

E = 2400

Electric Field Strength = 2400 Volts/meter

3 0

3 years ago

Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from

In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 $m/s^2$

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

$v = u + gt$

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

$49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs$

The arrow was in flight for 3.8 secs

6 0

3 years ago

A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo

m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

p₀ = m1 v₀₁

Moment after shock

$p_{f}$ = (m1 + m2) $v_{f}$

p₀ = $p_{f}$

m1 v₀₁ = (m1 + m2) $v_{f}$

$v_{f}$ = v₀₁ m1 / (m1 + m2)

$v_{f}$ = 4.4 600 / (600 + 500)

$v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

Em₀ = K = ½ (m1 + m2) $v_{f}$ ²

After compressing the spring

$E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

Em₀ = $E_{mf}$

½ (m1 + m2) $v_{f}$ ² = = ½ k x²

x = $v_{f}$ √ (k / (m1 + m2))

x = 2.4 √ (11/3000)

x = 0.396 m

7 0

3 years ago

How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),

fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice = 37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂ .

Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0

4 years ago

What current is needed in the solenoid's wires?

marta [7]

Using Ampere's Law, the magnetic field produced inside this solenoid is given by
B = uo N I / h
where uo is the vacuum permeability, N is the number of turns in the solenoid and h is the length of the solenoid. Earth's magnetic field is around 50 microteslas in North America thus the current needed in the solenoid is
I = B h / (uo N) = (50 E-6 ) (4) / ((4 pi E-7)(6000) ) = 0.026 A
I = 26 mA
So you need a current of around 26 mA.

5 0

3 years ago