Yo pls help me I'm desperate lowkey

Why would you have more gravitational force with Jupiter than the earth?

FromTheMoon [43]

Answer:

Jupiter has more mass

Explanation:

3 0

3 years ago

In a laboratory, you determine that the density of a certain solid is 5.23× 10 −6 kg/m m 3 . convert this density into kilograms

tamaranim1 [39]

Density is given as

$\rho = 5.23 * 10^{-6} kg/mm^3$

now we have to convert this density into $kg/m^3$

now we have

$1 m = 1000 mm$

$\rho = 5.23 * 10^{-6} \frac{kg}{(1*10^-3 m)^3}$

$\rho = 5.23 * 10^{-6} \frac{1*10^9 kg}{m^3}$

$\rho = 5.23 * 10^3 kg/m^3$

$\rho = 5230 kg/m^3$

8 0

3 years ago

How many moles of MgCI2 are there in 347 g of the compound

aliina [53]

Try dividing 347 by the molecular mass of Mg added to that of Cl2.

7 0

3 years ago

Identify and define characteristics of both visible and sound waves. These include but are not limited to frequency, pitch, ampl

jeyben [28]

<h3>Answer and explanation;</h3>

Sound waves are mechanical waves and are examples of the longitudinal type, a longitudinal wave involves the movement of molecules in the medium parallel to the direction of the wave.
Sound waves are longitudinal mechanical waves because they move particles parallel to their direction of travel. For example, if a sound wave travels from left to right.
Pitch is what we perceive as the low or high note quality of a sound. Pitch is determined by the frequency of the sound wave that moved through the air to our eardrums. Frequency is the time taken for a complete oscillation of a wave.
The more energy or work used to create the vibration, which in turn creates the sound wave, the greater the amplitude of the wave and the louder the sound. Wavelength is the distance between two successive rarefactions or compression in a longitudinal wave.
For example, if you were to lightly hit a drum with a drum stick, there would be a softer sound as opposed to the loud boom that would be created if you hit the drum with all of your might

3 0

3 years ago

1. You hit your hand on a desk and it doesn't hurt that much. The next day you

Usimov [2.4K]

Answer:

Problem 1: because, by the third law of Newton, the second time, the reaction force of the desk on the hand is stronger.

Problem 2: See the figure attached

Problem 3: See the figure attached

Problem 4: 8N

Problem 5: 10N

Problem 6: 0.0033 m/s²

Problem 7: 0.39 m/s² backwards

Explanation:

1. You hit your hand on a desk and it doesn't hurt that much. The next day you are carrying a heavy backpack in your hand and you hit the same desk. This time your hand hurts a lot when you hit the desk. Explain why your hand hurt the second time using concepts discussed in class.

When you hit the desk the first day, you were able to control the force with which you did it, thus you exerted a low force on the desk. By the third law of Newton, law of action and reaction, the force that the desk exerted on your hand is the same that your hand exerted on the desk, a low force.

The next day, the force with which you hit the desk is stronger because the heavy backpack pulled your hand toward the desk, thus you exerted a stronger force on the desk than the day before and, by the third law of Newton, the force that the desk exerted on you hand is also stronger.

In conclusion, your hand received a stronger hit back from the desk.

2. You are holding an apple over your head. Draw a free-body-diagram. A free-body-diagram is when you draw all the forces acting on individually objects that are free floating in space not touching anything else.

To draw the free-body-diagram, FBD, you replace the apple by a point, your hand by an arrow pointing upward, and the force that the Earth exerts on the apple (the weight of the apple) by an arrow pointing downward.

The apple is at rest on your hand, meaning that the two forces, the force of gravity and the force of your hand are balanced, making the net force zero. Thus, the size of the two arrows should be equal.

In the figure attached, the arrow pointing upward is labled H, and the arrow pointing downward Fg, for force of gravity.

See the FBD attached.

3. After a while, your hand/arm gets tired so you let go of the apple. Now the apple is falling through the air. Draw the free-body-diagram (and all the forces that are acting on the apple now).

After you let go of the apple, the apple starts falling through the air, at first the only force acting on the apple is the force of gravity, but at it moves down the air exerts a drag force. Many times this force is neglected, but that depends on the conditions of the problem.

For an apple falling down from your hand the drag force is usually neglected and it is said that the apple is in free fall.

Since, the problem states to include all the forces acting on the apple, your FBD must include the drag force, as an arrow pointing upward. This time the arrow pointing upward is shorter than the arrow pointing downward, because the drag force is less than the gravity force.

See the second figure attached.

4. If the apple has a mass of 0.8 kg, calculate the force due to gravity of the apple while it is falling through the air.

The magnitude of the force due to gravity on an object is equal to the product of the mass of the object and the acceleration of gravity. This force is what the weight of an object is.

The calculations are:

Formula:

Fg = mass × acceleration

Compute:

Fg = 0.8kg × 9.8m/s² = 7.84 N

Thus, rounding to one significant figure the force due to gravity is 8 newtons and is the wieght of the apple.

5. A box with mass (m) = 2 kg) has an acceleration of 5 m/sec². Calculate the total force acting on the box.

This time you must use the second law of Newton.

The second law of Newton states that the net force acting on an object is equal to the product of the mass and the acceleration.

Equation:

Net force = mass × acceleration

Substitute and compute:

Net force = 2kg × 5m/s² = 10N

Thus, the magnitud of the total force is 10 newtons.

6. Suppose that an astronaut outside pushes on a spaceship. The astronaut creates a force of 36N. If the mass of the spaceship is ms = 11,000kg, calculate the acceleration of the spaceship.

The physical law that rules this problem is the second law of Newton.

Net force = mass × acceleration

You must solve for the acceleration:

Acceleration = Net force / mass

Substitute and compute:

Acceleration = 36N / 11,000kg = 0.0033 m/s²

7. The astronaut forgot about Newton's 3rd law. The moment they push on the spaceship, they start to accelerate backwards. If the astronaut has a mass of ma = 92 kg, calculate the acceleration of the astronaut using information from the previous problem.

The third law of Newton is the law of action and reaction. The force that the astronaut exert on the spaceship is equal in magnitude to the force the spaceship exterts on the astronaut but backwards.

Then, using the second law of Newton with the same force and the mass of the astronaut you can calculate the acceleration of the astronaut.

acceleration = net force / mass

acceleration = 36N / 92kg = 0.39m/s² backwards

8 0

3 years ago