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3 years ago

If a car used 260.000 W of power to complete a race in 15 s. how much work did the car do?

Physics

1 answer:

marta [7]3 years ago

3 0

Work done by the car is 3900 J

Explanation:

Power and work are related by the equation, Power = Work Done/Time
Power is the rate at which work is done.
Here, the car uses power of 260 W and time taken is 15 s.

Work Done = Power × Time

= 260 × 15 = 3900 J

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A 0.5 m3 container is filled with a fluid whose specific volume is 0.001 m3/kg. At standard gravitational acceleration, the cont

xenn [34]

Answer: the contents of this container weighs 4905 kg.m/s²

Explanation:

Given that;

volume of a container V = 0.5 m³

we know that standard gravitational acceleration g = 9.81 m/s²

specific volume of liquid filled in the container v = 0.001 m³/kg

now we express the equation for weight of the container.

W = mg

W = (pV)g

W = Vg / ν

so we substitute

W = (0.5 m³)(9.81 m/s ) / 0.001 m³/kg

W = 4.905 / 0.001

W = 4905 kg.m/s²

Therefore, the contents of this container weighs 4905 kg.m/s²

5 0

2 years ago

What is the SI unit for momentum?

serg [7]

It’s a vector quantity, which means it possesses both magnitude and direction. So the SI unit would be B)kg•m/s

4 0

2 years ago

Read 2 more answers

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

3 years ago

A 5.31 kg object is swung in a vertical circular path on a string 2.99 m long. The acceleration of gravity is 9.8 m/s 2 . If the

11111nata11111 [884]

Answer:

T = 120.3 N

Explanation:

Since, the tension in the rope is acting against both the centripetal force and the weight of the stone. As both act downward towards center of the circle and tension acts towards point of support that is upward. So, tension will be equal to the sum of centripetal force and weight of the stone:

Tension = Centripetal Force + Weight of Stone

T = mv²/r + mg

where,

m = mass of stone = 5.31 kg

r = radius of circle = length of string = 2.99 m

g = 9.8 m/s²

Therefore,

T = (5.31 kg)(6.2 m/s)²/(2.99 m) + (5.31 kg)(9.8 m/s²)

T = 68.27 N + 52.03 N

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2 years ago

An imaginary line perpendicular to a reflecting surface is called ____refrac_____.

umka2103 [35]

Not totally sure but i would say a normal? its not refraction or incidence if its perpendicular and i dont think its a mirror if its an imaginary line so yeah normal (normals are always perpendicular to their surface too i think so)

4 0

3 years ago