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3 years ago

Calculate the normal force of a 1,200 kg boulder resting on the ground

Physics

2 answers:

Olin [163]3 years ago

4 0

Answer:

The normal force points upward and is 11,760 N

Explanation:

Resting on the ground = upward

1200 * 9.8 = 11, 760

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Darya [45]3 years ago

3 0

Answer:

11760N

Explanation:If the boulder is at rest and not going up or down we would take the 1200 kg and multiply it by 9.8 because

(sum)F = ma and since a is 0 the equation goes to

Fn - mg= 0 an that simplifies to

Fn=mg

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The relationship between an object's acceleration and its mass

Shalnov [3]

Answer:

Newton's second law

Explanation:

The relationship between mass and acceleration is described in Newton's Second Law of Motion. His Second Law states that the more mass an object has, more force is necessary for it to accelerate.

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3 years ago

The earthquake killed about 60 people and was the strongest earthquake to hit

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3 years ago

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Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

Levart [38]

Explanation:

Given that,

Charge 1, $q_1=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=4.39\times 10^{-6}\ C$

Distance between charges, r = 0.0209 m

1. The electric force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}$

F = -492.95 N

2. Distance between two identical charges, $r=0.0209\ m$

Electric force is given by :

$F=\dfrac{kq_3^2}{r^2}$

$q_3=\sqrt{\dfrac{Fr^2}{k}}$

$q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}$

$q_3=4.89\times 10^{-6}\ C$

Hence, this is the required solution.

3 0

2 years ago

A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

3 0

2 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So