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3 years ago

14

Suppose that two objects attract each other with a gravitational force of 18 units. If the distance between the two objects is t

ripled, then what is the new force of attraction between the two objects?

1 answer:

Burka [1]3 years ago

4 0

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between the 2 objects.

now the distance is tripled. that means

Fgravitynew = G*(mass1*mass2)/(3D)² =

= G*(mass1*mass2)/(9D²) =

= (G*(mass1*mass2)/D²) / 9 = Fgravity/9

so, the new force of attraction is 18/9 = 2 units.

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Answer:

Wave A

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3 years ago

a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio

Harman [31]

Answer:

The angular acceleration is $\alpha = 3.235 \ rad/s ^2$

Explanation:

From the question we are told that

The moment of inertia is $I = 0.034\ kg \cdot m^2$

The net torque is $\tau = 0.11\ N \cdot m$

Generally the net torque is mathematically represented as

$\tau = I * \alpha$

Where $\alpha$ is the angular acceleration so

$\alpha = \frac{\tau }{I}$

substituting values

$\alpha = \frac{0.1 1}{ 0.034}$

$\alpha = 3.235 \ rad/s ^2$

6 0

3 years ago

A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?

Mashcka [7]

Answer:

<em>v=40 m/s south</em>

Explanation:

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It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

$p=m.v$

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

$\displaystyle v=\frac{p}{m}$

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

$\displaystyle v=\frac{6}{0.15}=40\ m/s$

The velocity is directed to the south

3 0

4 years ago

The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh

jeyben [28]

Explanation:

Given that,

Electrostatic force, $F=3.7\times 10^{-9}\ N$

Distance, $r=5\ A=5\times 10^{-10}\ m$

(a) $F=\dfrac{kq^2}{r^2}$ , q is the charge on the ion

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

$q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}$

$q=3.2\times 10^{-19}\ C$

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

$n=\dfrac{q}{e}$

$n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}$

n = 2

Hence, this is the required solution.

8 0

4 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

4 years ago