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3 years ago

Solve the equation 3x^2+8x+5=0

Physics

1 answer:

Pachacha [2.7K]3 years ago

4 0

$3x^2+8x+5=0\\
3x^2+3x+5x+5=0\\
3x(x+1)+5(x+1)=0\\
(3x+5)(x+1)=0\\
x=-\dfrac{5}{3} \vee x=-1$

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Answer:

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3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

MAVERICK [17]

Answer:

t₂ = 3.89 s

Explanation:

given,

speed of car = 23 m/s

speed of motorcycle = 23 m/s

after time of 4 s distance between them is equal to = 53 m

motorcycle accelerates at = 7 m/s

time taken to catch up with car = ?

let t₂ be the time in which motorcycle catches car.

distance traveled by car in t₂ s

d = 23 t₂ + 53

distance traveled by motorcycle

using equation of motion

$s = ut + \dfrac{1}{2}at^2$

$s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

now, equating both the distances

$23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

$t_2^2 = 15.143$

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time taken by the motorcycle to catch the car is equal to 3.89 s

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3 years ago

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the

german

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of $u=19\ m/s$

When the hand is $h=2.5\ m$ above the ground

Using the equation of motion we can write

$h=ut+\dfrac{1}{2}at^2$

Substitute the values

$\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]$

Thus, the ball will take 4 s to hit the ground.

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3 years ago

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Ann [662]

Answer:

Explanation:

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3 years ago

Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an

Paladinen [302]

Answer:

$\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}$

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F=m \times a$

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

m= 8 kg
a= 1.6 m/s²

Substitute these values into the formula.

$F= 8 \ kg * 1.6 \ m/s^2$

Multiply.

$F= 12.8 \ kg*m/s^2$

<h3>2. Acceleration of the 2nd Object </h3>

Now, use the force we just calculated to complete the second part of the problem. We use the same formula:

$F= m \times a$

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

F= 12.8 kg *m/s²
m= 2 kg

Substitute the values into the formula.

$12.8 \ kg*m/s^2= 2 \ kg *a$

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

$\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}$

$\frac {12.8 \ kg*m/s^2}{2 \ kg}=a$

The units of kilograms cancel.

$\frac {12.8}{2}\ m/s^2=a$

$6.4 \ m/s^2=a$

The acceleration is 6.4 meters per square second.

4 0

2 years ago