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mel-nik [20]
3 years ago
6

During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. What was the ball's speed 2.2 s after she kicked

it?
Physics
1 answer:
ki77a [65]3 years ago
5 0

<u>Answer</u>

3.44 m/s


<u>Explanation</u>

The motion apply the equations of Newton's law of motion. The ball is acceleration is -9.8 m/s² (acceleration due to gravity. It is negative because the ball is going against gravity, so it is decelerating).

The first equation of Newton's law of motion is;

V = U + at

Where V is the final velocity, U is the initial velocity, a is acceleration and t the time taken.

V = 25 + (-9.8 × 2.2)

    = 25 - 21.56

     = 3.44 m/s

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4 0
2 years ago
The work done to compress a spring with a force constant of 290.0 N/m a total of 12.3 mm is: a) 3.57 J b) 1.78 J c) 0.0219 J d)
iren2701 [21]

Answer:

Work done, W = 0.0219 J

Explanation:

Given that,

Force constant of the spring, k = 290 N/m

Compression in the spring, x = 12.3 mm = 0.0123 m

We need to find the work done to compress a spring. The work done in this way is given by :

W=\dfrac{1}{2}kx^2

W=\dfrac{1}{2}\times 290\times (0.0123)^2

W = 0.0219 J

So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.

7 0
3 years ago
A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the
podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

F_x = Fcos30


F_y = Fsin30


Now the normal force on the block is given as

N = Fsin30 + mg

N = 0.5F + (18\times 9.8)

N = 0.5F + 176.4

now the friction force on the cart is given as

F_f = \mu N

F_f = 0.625(0.5F + 176.4)

F_f = 110.25 + 0.3125F

now if cart moves with constant speed then net force on cart must be zero

so now we have

F_f + F_x = 0

Fcos30 - (110.25 + 0.3125F) = 0

0.866F - 0.3125F = 110.25

F = 199.2 N

so the force must be 199.2 N

8 0
3 years ago
Why is the scientific method described as cyclic?
Troyanec [42]
Because the scientific method can go around in a circle as many times as neccisary to get the results you need
8 0
3 years ago
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