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viktelen [127]
3 years ago
13

What does a force acting on an object do to that object';s motion?

Physics
1 answer:
ch4aika [34]3 years ago
8 0
Accourding to newtons second law of motion:
Force = mass * acceleration
F = ma
a = F/m
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If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the tra
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If the distance between a neutral atom and a point charge is quadrupled, by what factor does the force on the atom by the point
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A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
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Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
4 years ago
A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k
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Answer:

v_{f,w} = 1.791\,\frac{m}{s}, v_{f,c} = 0.972\,\frac{m}{s}

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}

The final angular speed is:

\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}

\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev}  \right)

\omega_{f} \approx 5.116\,\frac{rad}{s}

The tangential velocities of the wheel and the clay are, respectively:

v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )

v_{f,w} = 1.791\,\frac{m}{s}

v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )

v_{f,c} = 0.972\,\frac{m}{s}

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