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3 years ago

when an object moves down and does not stop which force is acting more strongly on the object, friction or gravity? explain

Physics

1 answer:

lina2011 [118]3 years ago

7 0

Gravity acts more strongly on the object.

<u>Explanation:</u>

When an object is dropped from a height, it reaches the ground despite friction acting on it because the force of gravity acting on it is stronger than the air resistance and friction. Air resistance and friction acts upward and prevents the ball from falling. However, it is negligible. The gravity acting on the object is so strong that it pulls the object towards earth with a constant acceleration called as acceleration due to gravity which has a constant value of 9.8m/s².

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mafiozo [28]

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

Magnitude of first vector = $|A| = 13\ \text{units}$

Angle = $\theta_1=27^{\circ}$

Magnitude of second vector = $|B| = 11\ \text{units}$

Angle = $\theta_2=45^{\circ}$

x component of first vector

$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

y component of first vector

$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

x component of second vector

$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

y component of first vector

$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

$C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}$

Magnitude of the sum of the vectors would be

$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

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