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goblinko [34]
4 years ago
14

The universal law of gravitation states that the force of attraction between two objects depends on which quantities? the masses

of the objects and their densities the distance between the objects and their shapes the densities of the objects and their shapes the masses of the objects and the distance between them
Physics
2 answers:
LiRa [457]4 years ago
6 0

Answer : The gravitational force depends on the the masses of the objects and the distance between them.

Explanation :

The universal law of gravitation gives the relationship between the masses and the distance between them.  

It states that there exists a force in the universe which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, it can be written as :

F=G\dfrac{m_1m_2}{d^2}          

Where,

G is the universal constant, G = 6.67 × 10⁻¹¹ N.m²/Kg²

m₁ and m₂ are the mass

d is the distance between them

So, the correct statement is " the masses of the objects and the distance between them".

Vedmedyk [2.9K]4 years ago
3 0
The mass of objects and distance between them
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Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the b
Murljashka [212]

Answer:

a_total = 14.022 m/s²

Explanation:

The total acceleration of a uniform circular motion is given by the following formula:

a=\sqrt{a_c^2+a_T^2}         (1)

ac: centripetal acceleration

aT: tangential acceleration

Then, you first calculate the centripetal acceleration by using the following formula:    

a_c=r\omega^2

r: radius of the circular trajectory = 2.0m

w: final angular velocity  of the ball = 7.0 rad/s

a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}        

Next, you calculate the tangential acceleration. aT is calculate by using:

a_T=r\alpha    (2)

α: angular acceleration

The angular acceleration is:

\alpha=\frac{\omega_o-\omega}{t}

wo: initial angular velocity = 13 rad/s

t: time = 15 s

Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:

a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}

Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:

a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}

The total acceleration of the ball is 14.022 m/s²

5 0
3 years ago
Personality theorists often focus on _____________.
lubasha [3.4K]
A. types or traits is the answer
8 0
4 years ago
A block of weight 45.7 N is hanging from a rope. The tension from the rope is pulling upward on the block. The block is accelera
viktelen [127]
<h2>Answer:</h2><h2></h2>

52.555 N

<h2>Explanation:</h2>

Let's use Newton's second law of motion here which states that the resultant force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) due to this force. i.e

∑F = m x a            ---------------------(i)

<em>Now, let's get the resultant force;</em>

Two main forces are acting on the rope;

i. the weight (W) of the block acting downwards.

Where;

W = mass of block(m) x gravity(g) = m x g

ii. the tension (T) in the rope acting upwards.

Therefore, the resultant force is the vector sum of these two forces as follows;

∑F = - W + T            [upward motion is taken as positive. hence -W and +T]

<em>Substitute ∑F = - W + T into equation (i) as follows;</em>

- W + T = m x a      ---------------------(ii)

<em>From the question;</em>

* Weight (W) of the block = 45.7N

=> mass (m) of the block = W / g = 45.7 / 10               [Taking g = 10m/s²]

=> m = 4.57 kg

* acceleration (a) = 1.50m/s²

<em>Substitute these values into equation (ii) as follows;</em>

- 45.7 + T = 4.57 x 1.50

- 45.7 + T = 6.855

<em>Solve for T;</em>

T = 6.855 + 45.7

T = 52.555 N

Therefore, the tension in the rope is 52.555 N

   

6 0
4 years ago
The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.
Lesechka [4]

The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it. Because the contact between the bolt and the   pliers working surface is less.

<h3>What is mechanical advantage ?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of the given mechanical machine.

The efficiency to open the stubborn bolt depends upon the contact between the working surface of the pliers and the bolt.

The contact between the bolt and the top pair of pliers working surface is less. Its mechanical advantage is less.

Hence, the top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.

To learn more about the mechanical advantage, refer to the link;

brainly.com/question/7638820

#SPJ1

3 0
2 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 14 s until its motor stops. Disreg
telo118 [61]

Answer:

The maximum height of the rocket will be 1.0 × 10⁴ m.

Explanation:

Hi there!

The height of the rocket at time "t" can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²  (when the rocket has an upward acceleration)

y = y0 + v0 · t + 1/2 · g · t²  (after the motor of the rocket stops)

Where:

y = height.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to the motors.

g = acceleration due to gravity.

The velocity of the rocket can be calculated as follows:

v = v0 + a · t  (while the motor is running)

v = v0 + g · t  (after the motor stops)

Where "v" is the velocity of the rocket at time "t".

The rocket rises with upward acceleration for 14 s. After that, the rocket starts being accelerated in the downward direction due to gravity. But it will continue going up after the motor stops because the rocket has initially an upward velocity that will be reduced until it becomes zero and the rocket starts to fall.

Let´s find the height reached by the rocket while it was accelerated in the upward direction (the origin of the frame of reference is located at the launching point and upward is the positive direction):

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · 14 s + 1/2 · 28 m/s² · (14 s)²

y = 2.7 × 10³ m

Now let´s find the velocity reached in that time:

v = v0 + a · t

v = 28 m/s² ·14 s

v = 3.9 × 10² m/s

Now, let´s find the maximum height reached by the rocket using the equations of height and velocity after the motor stops:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Notice that now v0 and y0 will be the velocity and height reached while the rocket was being accelerated in the upward direction, respectively.

Let´s find at which time the rocket reaches its maximum height. With that time, we can calculate the max-height.

At the maximum height, the velocity of the rocket is zero, then:

v = v0 + g · t

0 = 3.9 × 10² m/s - 9.8 m/s² · t

-3.9 × 10² m/s/ -9.8 m/s² = t

t = 40 s

After the motor stops, it takes the rocket 40 s s to reach the maximum height.

Using the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 2.7 × 10³ m +  3.9 × 10² m/s · 40 s - 1/2 · 9.8 m/s² · (40 s)²

y = 1.0 × 10⁴ m

The maximum height of the rocket will be 1.0 × 10⁴ m

4 0
3 years ago
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