PHYSICS HIGH SCHOOL

After tasting the salt in the ocean water on a visit to the beach one day, you recall from class that the salt is the __________ in the ocean water. A. solid
B. solute
C. solution
D. solvent

Answers

Answer 1
Answer: Solid is a stateof matter having no space in a volume, made entirely of compact material.
Solute is adissolved substance is a dissolved substance in a solution.Solution is theone that dissolves the solute.
Solvent is theone that dissolves either a solution or solute.
Based on thedefinitions, the answer is letter B. solute. 
Answer 2
Answer:

Answer:

B. Solute

Explanation:

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The closer together a magnets field lines are,

Answers

the stronger they will be

Compare the time taken by the RC car to cover the same distance. Do some go faster or slower?

Answers

There isn;t enough given values to define and determine what is asked. But since
distance = speed / time, time = distance / speed


Calculate the acceleration of a boat that has been cruising at 15 m/s but speeds up to 45 m/s in about 6 seconds. Don’t forget your units, m/s2. In order to calculate the acceleration of the boat we first have to identify the suited formula for acceleration in this scenario. 

Hence, a = vf – vi / t. Vi = 15 m/s Vf = 45 m/s T = 6 seconds   

Solution: 
A = 45 m/s – 15m/s / 6s 
A = 5m/s^2

The shorter the time required to cover the same distance, the greater the speed of the RC car.

Further explanation

Acceleration is rate of change of velocity.

\large {\boxed {a = (v - u)/(t) } }

\large {\boxed {d = (v + u)/(2)~t } }

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

The formula for finding speed is:

\large {\boxed {v = (s)/(t)} }

v = velocity (m/s)

s = distance (m)

t = time taken (s)

From the formula above, it is clear that speed is inversely proportional to travel time. The shorter the time required to cover the same distance, the greater the speed of the RC car.

For example, let's look at the data in the table in the attachment. RC Car B has the greatest speed

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

a missile is moving 1350 m/s at a 25.0 deg angle. it needs to hit a target 23,500 m away in 55.0 deg direction in 10.20 s. what is the magnitude of its final velocity?

Answers

In this question we have given

velocity of missile=1350m/s

angle at which missile is moving=25degree

distance between missile and targets=23500m

angle between target and missile=55degree

time=10.2s

To find the final velocity of missile we will first find the acceleration required

Let x be the horizontal component of distance

x - vertical component of distance

t-time

ax- horizontal component of acceleration

ay-Vertical component of acceleration

Vx-horizontal component of velocity

Vy-Vertical component of velocity


horizontally: x = Vx*t + ½*ax*t²  

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²  

ax = 19.2 m/s²  

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s  

similarly vertically:

y = Vy*t + ½*ay*t² 

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²  

ay = 258 m/s²  


V'y = Vy + ay*t = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s  

Therefore

V = √(V'x² + V'y²) = 3504 m/s  

therefore magnitude of final velocity of missile=3504m/s

Which is not true about neutron? (a) It is neutral particle (b) It is highly unstable in nature (c) It is present in the nucleus. (d) It contributes to the most of the atom​

Answers

(b) It is highly unstable in nature. Neutrons are actually quite stable and are present in the nucleus.

A heavier mass m1 and a lighter mass m2 are 20.5 cm apart and experience a gravitational force of attraction that is 8.50 10-9 N in magnitude. The two masses have a combined value of 5.80 kg. Determine the value of each individual mass.m1 = ____ kgm2 = ____ kg

Answers

Final answer:

The gravitational force between two individual masses can be calculated using Newton's law of universal gravitation. Given the total mass and the gravitational force, we can derive a quadratic equation to solve for the individual masses. The quadratic formula is then used to solve this equation.

Explanation:

The properties of the gravitational force between two masses can be derived from Newton's law of universal gravitation: F = G * (m1*m2) / r². In this equation, F is the force between the masses, G is the gravitational constant (approximately 6.674 * 10^{-11} N*(m²/kg²)), m1 and m2 are the masses, and r is the distance between the centers of the two masses.

Given that F = 8.50 * 10^{-9} N, r = 0.205 m (we need to convert cm to m), and m1 + m2 = 5.80 kg, we can substitute the known values into the equation to solve for the two masses.

This results in a quadratic equation in terms of one of the masses, which can be solved using the quadratic formula: -b ± sqrt(b² - 4ac) / 2a. Once we solve this equation we get the individual values of m1 and m2.

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Final answer:

To find the values of m1 and m2, we can use the formula for gravitational force and solve for each mass. By substituting the given values, solving the equation, and simplifying, we can find the individual masses. The value of m1 is approximately 0.07415 kg and the value of m2 is approximately 0.05728 kg.

Explanation:

To determine the value of the individual masses, we can use the formula for gravitational force: F = G * (m1 * m2) / r^2, where G is the gravitational constant and r is the distance between the masses. Rearranging the formula, m1 = (F * r^2) / (G * m2). Substituting the given values, we can solve for m1 and m2.

  1. Convert the distance of 20.5 cm to meters by dividing by 100: 20.5 cm / 100 = 0.205 m.
  2. Substitute the given values into the formula: m1 = (8.50 * 10^(-9) N * (0.205 m)^2) / (6.67 * 10^(-11) N m^2/kg^2 * m2).
  3. Substitute the total mass of 5.80 kg into the equation: m1 = (8.50 * 10^(-9) N * (0.205 m)^2) / (6.67 * 10^(-11) N m^2/kg^2 * (5.80 kg - m1)).
  4. Simplify the equation by multiplying and dividing: m1 = (3.455 * 10^(-9) kg m^2 / s^2) / (6.67 * 10^(-11) m^3 kg^(-1) s^(-2) - 3.455 * 10^(-9) kg m^2 / s^2).
  5. Multiply both sides by the denominator and simplify: (6.67 * 10^(-11) m^3 kg^(-1) s^(-2) - 3.455 * 10^(-9) kg m^2 / s^2) * m1 = 3.455 * 10^(-9) kg m^2 / s^2.
  6. Distribute and combine like terms: (6.67 * 10^(-11) m^3 kg^(-1) s^(-2)) * m1 - (3.455 * 10^(-9) kg m^2 / s^2) * m1 = 3.455 * 10^(-9) kg m^2 / s^2.
  7. Combine the like terms and isolate m1: (6.67 * 10^(-11) m^3 kg^(-1) s^(-2) - 3.455 * 10^(-9) kg m^2 / s^2) * m1 = 3.455 * 10^(-9) kg m^2 / s^2.
  8. Divide both sides by the remaining term: m1 = (3.455 * 10^(-9) kg m^2 / s^2) / (6.67 * 10^(-11) m^3 kg^(-1) s^(-2) - 3.455 * 10^(-9) kg m^2 / s^2).
  9. Use a calculator to solve for m1: m1 ≈ 7.415 * 10^(-2) kg.
  10. Substitute the value of m1 into the equation for the total mass: 5.80 kg = m1 + m2.
  11. Solve for m2: m2 = 5.80 kg - m1.
  12. Substitute the value of m1: m2 ≈ 5.80 kg - 7.415 * 10^(-2) kg.
  13. Use a calculator to solve for m2: m2 ≈ 5.728 * 10^(-2) kg.

Therefore, the value of m1 is approximately 0.07415 kg and the value of m2 is approximately 0.05728 kg.

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An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outdoor temperature (and the gasoline in a tanker truck) is 97.0°F, how many gallons from the truck can be poured into the tank? Assume the temperature of the gasoline quickly cools from 97.0°F to 52.0°F upon entering the tank. (The coefficient of volume expansion for gasoline is 9.6 10-4 (°C)−1.)

Answers

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ)  β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

    = 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

    = 1.07 × 10³(1 - [0.024]²)

    =  1.07 × 10³(1 - 0.000576)

    = 1.07 × 10³(0.999424)

    = 1069.38 gallons

Final answer:

To calculate the amount of gasoline that can be poured into the tank, we need to find the change in volume of the gasoline when its temperature changes from 97.0°F to 52.0°F. Using the equation for volume expansion, we can calculate this change in volume to be approximately 258 gallons.

Explanation:

To calculate the amount of gasoline that can be poured into the tank, we need to find the change in volume of the gasoline when its temperature changes from 97.0°F to 52.0°F. We can use the equation for volume expansion to calculate this change in volume:

ΔV = V₀ * β * ΔT

Where ΔV is the change in volume, V₀ is the initial volume, β is the coefficient of volume expansion, and ΔT is the change in temperature.

In this case, the initial volume V₀ is 1.07 * 10³ gallons, the coefficient of volume expansion β is 9.6 * 10⁻⁴ (°C)⁻¹, and the change in temperature ΔT is (52.0°F - 97.0°F) = -45.0°F.

Converting the change in temperature to Celsius: ΔT = (45.0°F) * (5/9) = -25.0°C.

Plugging in these values into the equation, we get:

ΔV = 1.07 * 10³ * 9.6 * 10⁻⁴ * -25.0 = -258 gallons.

Therefore, when the gasoline is poured into the tank, approximately 258 gallons will be poured out of the truck.

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