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3 years ago

Two 6.0 kg masses are separated by a distance of 4.0 m. What is the magnitude of the gravitational force between the two masses?

Physics

1 answer:

gogolik [260]3 years ago

5 0

Answer: $1.50075*10^{-10}\ N$

Explanation:

For this exercise we need to use the Gravitational Force Formula. This is:

$F_g=G(\frac{m_1*m_2}{r^2})$

Where $F_g$ is the gravitational force between two objects, $G$ is the gravitational constant ( $G=6.67*10^{-11}\ \frac{Nm^2}{kg^2}$ ) and $r$ is the distance between the objects.

We know that:

$m_1=6.0\ kg\\\\m_2=6.0\ kg\\\\r=4.0\ m$

Therefore, in order to calculate the magnitude of the gravitational force between the two masses, we must substitute these values into the formula.

Then we get:

$F_g=(G=6.67*10^{-11}\ \frac{Nm^2}{kg^2})(\frac{(6.0\ kg)(6.0\ kg)}{(4.0\ m)^2})\\\\F_g=1.50075*10^{-10}\ N$

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A shot is fired at an angle of 60 degree horizontal with Kinetic energy E. If air resistance is ignored, the K.E at the top of t

Lapatulllka [165]

I'm not sure what "60 degree horizontal" means.

I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.

Now, I'll answer the question that I have invented.

When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)

-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.

-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.

-- So at the top of its trajectory, its KE is 0.25 of what it had originally.

That's E/4 .

3 0

3 years ago

Solenoid 2 has twice the radius and six times the number of turns per unit length as solenoid 1. The ratio of the magnetic field

Xelga [282]

Answer:

Explanation:

The magnetic field inside a solenoid is given by the following formula:

$B = \mu_{0}nI$

where,

B = Magnetic Field Inside Solenoid

μ₀ = permittivity of free space

n = No. of turns per unit length

I = Current Passing through Solenoid

For Solenoid 1:

$B_{1} = \mu_{0}n_{1}I ------------------- equation 1$

For Solenoid 2:

n₂ = 6n₁

Therefore,

$B_{1} = \mu_{0}n_{2}I\\B_{1} = 6\mu_{0}n_{1}I ----------------- equation 2$

Diving equation 1 and equation 2:

$\frac{B_{2}}{B_{1}} = \frac{6\mu_{0}nI}{\mu_{0}nI}\\\\\frac{B_{2}}{B_{1}} = 6$

Hence, the correct option is:

8 0

3 years ago

A skier accelerates down the hill at 3m/s2 how fast is he going in 4 seconds

almond37 [142]

Answer:

Explanation: simple kinematics

we suppose that initially vo= 0 so if the skier moves 4s :

vf = vo +at = 0 + 3*4 = 12 m/s

best wishes from colombia

7 0

3 years ago

The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla

kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

$v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\$

To find the time we must use another kinematic equation.

$v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]$

7 0

3 years ago

18 points! Brainliest! Physics!

Naddik [55]

You will use the Pythagorean Theorem to solve it.
c^2 = a^2 + b^2
c^2 = (1.5)^2 + (2)^2
c^2 = 6.25
c = square root of 6.25
c = 2.5
I hope this helps!

3 0

4 years ago