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4 years ago

A race car accelerates from rest to a velocity of +90 m/s over a distance of 423m. Determine the acceleration of the race car.

Physics

1 answer:

dlinn [17]4 years ago

4 0

Answer:

9.57m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity = 90m/s

Distance = 423m

Unknown:

Acceleration of the race car = ?

Solution:

To solve this problem, we should apply one of the appropriate motion equations;

V² = U² + 2as

Where V is the final velocity

U is the initial velocity

a is the acceleration

s is the distance

90² = 0² + 2 x a x 423

8100 = 846a

a = 9.57m/s²

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2. A 56N force is exerted on an 8kg object. What is the object's acceleration?

White raven [17]

acceleration = 7 m/s²

Explanation:

To determine the object's acceleration we use the following formula:

force (N) = mass (kg) × acceleration (m/s²)

acceleration (m/s²) = force (N) / mass (kg)

acceleration = 56 N / 8 kg

acceleration = 7 m/s²

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acceleration

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3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

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6 0

3 years ago

A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t

maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and $\theta$ is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

$y=vt+\frac{1}{2}at^2$

where $y=vertical\ distance$

$a=acceleration$

$t=time$

$v=initial\ velocity$

here initial velocity is v\sin \theta [/tex] so

$y=v\sin \theta \times 1-\frac{1}{2}gt^2$

$y=v\sin \theta -0.5g$

In absence of gravity

$y_2=v\sin \theta \times t$

$y_2=v\sin \theta \times 1$

$\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m$

4 0

3 years ago

Which of the following helps to explain how hydraulic lifts work?

Gnesinka [82]

Answer : C. Pascal's principle.

Explaination : Pascal's principle (well-known as Pascal's law) states that if a closed container contains a fluid at rest, then a small change in pressure at one side of the fluid is transmitted to each and every part of the fluid and also to the walls of the container without any loss. In a hydraulic lift, we need the same mechanism to work and so we take the help of Pascal's principle.

Hence, the correct option is C. Pascal's principle.

8 0

4 years ago

Read 2 more answers

Two spherical objects have masses of 6.2 x 105 kg and 13 x 103 kg. The gravitational attraction between them is 130 N. How far a

Neporo4naja [7]

Answer:

r = 6.4 cm

Explanation:

F = GMm/r²

r = √(GMm/F)

r = √((6.674e-11)(6.2e5)(13e3)/130)

r = 0.06432... m

Those are some high density materials!

8 0

3 years ago