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olasank [31]
2 years ago
14

At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball

's acceleration is 890 m/s2 and the vertical or y component of its acceleration is 1100 m/s2. The ball's mass is 0.41 kg. What is the magnitude of the net force acting on the soccer ball at this instant
Physics
1 answer:
mixas84 [53]2 years ago
4 0

Answer:

580.13N

Explanation:

The magnitude of the net force is expressed using the expression

F = ma

m is the mass

a is the net acceleration

Given

Mass = 0.41kg

Get the net acceleration

a = √(ax)²+(ay)²

a = √890²+1100²

a = √792100+1210000

a = √2002100

a = 1414.96m/s²

Get the required net force acting on the ball

F = ma

F = 0.41×1414.96

F = 580.13N

Hence the net force acting on the soccer ball at this instant is 580.13N

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Answer:

the answer is C

Explanation:

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A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
dimaraw [331]

Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

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How long would it take to go 11 miles at 22 mph?
harkovskaia [24]
It would tack about 3.2 h

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A vocalist with a bass voice can sing as low as 92 Hz.
Inessa05 [86]

Answer:

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icang [17]

Answer:

answer is 61.8 m/s; 32.9

l am not sure

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2 years ago
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