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GREYUIT [131]
3 years ago
9

A soccer ball is kicked with a velocity of 8 m/s at an angle of 23°. what is the ball's acceleration in the vertical direction o

f the flies through the air?
Physics
2 answers:
Aloiza [94]3 years ago
3 0
Regardless of the speed of the ball or its angle, once it has left the kickers foot it's acceleration is always g downward. -9.81m/s^2
Nata [24]3 years ago
3 0

Answer:

9.81 m/s^2 vertically downwards

Explanation:

When a body is falling freely or throwing vertically upwards, the acceleration acting on the body is equal to the acceleration due to gravity. It always acts in vertically downwards direction and it's value is 9.81 m/s^2.

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The mass of an atom of carbon-12 (Z = 6) is less than the mass of 6 protons and 6 neutrons because of
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Because 2 carbon attach to one's of each of the others
4 0
3 years ago
A golf ball has a mass of 46 grams, and when struck by a driver reaches a speed of 42 m/s in a time of 0.00050 seconds. what was
Fittoniya [83]
Impulse in physics is the integral of force, F, with respect with time, t. This value is a vector quantity since force is a vector quantity as well. It can be calculated from the product of force and time. We do as follows:

Impulse = Ft
               = m(a)(t)
               = m(v/t)t
               = 0.046 (42/0.0005) (0.0005)
               = 1.932 N-s
4 0
3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
An object of mass 0.5 kg is swung in uniform circular motion. The radius is 2 meters, and the force exerted is 4 N. Calculate th
Katarina [22]
Using
F= mv²/r
4 = 0.5×v² / 2
8 /0.5 = v²
v²=16
v= √16
v= 4 ms-¹
8 0
3 years ago
A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration o
Sergeeva-Olga [200]

Answer:

a=2.4\ m/s^2

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

d=ut+\dfrac{1}{2}at^2

or

d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2

So, the acceleration of the car is 2.4\ m/s^2.

5 0
3 years ago
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