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3 years ago

Please help correct answering get 20 points (physical science btw )

Physics

1 answer:

Zinaida [17]3 years ago

6 0

Answer:

Acceleration = 3.67 m/s²

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Given the following data;

Initial velocity, u = 5 m/s

Final velocity, v = 27

Time, t = 6

To find the acceleration;

Acceleration = (v - u)/t

Acceleration = (27 - 5)/6

Acceleration = 22/6

Acceleration = 3.67 m/s²

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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

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Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

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4 years ago

Which parameter of the present universe, more than any other, is considered to be critical in determining the ultimate fate of t

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The slope and curvature of space time which is being derived from the Einstein's law of gravitation which was modified later it gives three slopes value (-,0,+ )

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3 years ago

A piece of indium with a mass of 16.6 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to

blagie [28]

Answer:

the density of indium is 7.2 g/cm^3

Explanation:

The computation of the density of indium is shown below:

Given that

Mass = 16.6 g

Volume = 48.6 c,^3 - 46.3cm^3 = 2.3 cm^3

Based on the above information

As we know that

Density = mass ÷ volume

So,

= 16.6g ÷ 2.3 cm^3

= 7.2 g/cm^3

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We simply applied the above formula so that the correct value could come

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3 years ago

1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential

cupoosta [38]

<h3>Answer:</h3>

$\displaystyle U_s = 0.6 \ J$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: $\displaystyle U_s = \frac{1}{2} k \triangle x^2$

U is energy (in J)
k is spring constant (in N/m)
Δx is displacement from equilibrium (in m)

<h3>Explanation:</h3>

<u>Step 1: Define</u>

k = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

Substitute in variables [Elastic Potential Energy]: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2$
Evaluate exponents: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = 0.6 \ J$

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Answer:

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Explanation:

Below is an attachment containing the solution.

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