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g100num [7]
4 years ago
9

Which change will always result in an increase in gravitational force between two objects

Physics
2 answers:
grin007 [14]4 years ago
7 0
Increasing the masses of the objects and decreasing the distance between the objects
jolli1 [7]4 years ago
5 0

Answer: Hello, the gravitational force has the form F = G \frac{M1*M2}{r^{2} }, where M1 and M2 are the masses of the objects, G is a constant and r is the distance between the two objects.

first: increasing the mass of the objects, the masses are in the numerator, then if whe increase either M1 or M2 (or both of them) then the gravitational force will increase.

second: decreasing the distance between the objects ( r in this case). r is in the denominator, si if r is smaller, the result of the division will be bigger, so if r decrease, then the gravitational force increase.

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Complete the mechanism for the given decarboxylation by adding any missing atoms, bonds, charges, nonbonding electrons, and curv
AlexFokin [52]

The decarboxylation reaction involves the removal and replacement of a carboxyl group with hydrogen.

<h3>What is a decarboxylation reaction?</h3>

A decarboxylation reaction is one in which the carboxyl group in a carboxylic acid is replaced with hydrogen.

In living organisms, decarboxylation reactions are catalyzed by enzymes called decarboxylases or carboxy-lyases.

The decarboxylation of beta-carbonyl esters proceeds through a cyclic transition state giving an enol intermediate which then tautomerises to the carbonyl.

Therefore, a decarboxylation reaction involves the removal and replacement of a carboxyl group with hydrogen.

Leran more about decarboxylation at: brainly.com/question/893601

5 0
2 years ago
The ball is moving at a constant speed of 0.5 m/s for 2.3 seconds how far does it go?
yarga [219]

Distance = (speed) x (time)

Distance = (0.5 m/s) x (2.3 s)

Distance = (0.5 x 2.3) m

Distance = 1.15 meters

7 0
4 years ago
Can someone plzzzz help with these 3 I don’t understand
Aleks [24]

12.) Active transport because the cell must use energy to move large particles across the membrane.

13.) Photosynthesis takes place in plant leaves containing the chlorophyll pigment. Cellular respiration takes place in the cytoplasm and mitochondria of the cell. ... Cellular respiration uses glucose molecules and oxygen to produce ATP molecules and carbon dioxide as the by-product.

14.) In cells with a nucleus, as in eukaryotes, the cell cycle is also divided into two main stages: interphase and the mitotic (M) phase (including mitosis and cytokinesis). During interphase, the cell grows, accumulating nutrients needed for mitosis, and undergoes DNA replication preparing it for cell division.

4 0
3 years ago
Jim is driving a 2268-kg pickup truck at 19 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
lord [1]

Answer:

The initial kinetic energy of the truck is 409374 J

Explanation:

This problem can be solved in two ways. Let´s solve it first in the easiest way.

The kinetic energy is calculated using this equation:

E = 1/2 · m · v²

Where:

E = kinetic energy

m = mass

v = velocity

Then, the kinetic energy of the truck will be:

E = 1/2 · 2268 kg · (19 m/s)² = 409374 J

And that´s it.

But we can complicate it a bit:

The kinetic energy is the work needed to move an object from rest to a desired velocity. If the object is moving, the work needed to stop it must be of the same magnitude as its kinetic energy (in the opposite direction to the movement).

The equation for work is:

W = F · d

Where:

W= work

F = force

d = distance

We know the magnitude of the force applied to the truck, but we do not know for how much distance that force was applied. The distance can be calculated using the equation for the position of an object moving in a straight line:

x = x0 + v0 · t + 1/2 · a · t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

But we still do not know the time nor the acceleration.

The acceleration can be obtained from the equation of force:

F = m · a

Where

F = force

m = mass

a = acceleration

Then:

900 N = 2268 kg ·a

a = 900 N /2268 kg = 0.397 m/s²

Now, we can calculate the time needed for the truck to stop. We know that at the final time, the velocity is 0. Then, we can use the equation for velocity to obtain that time:

v = v0 + a · t

Where:

v = velocity at time t

v0 = initial velocity

a = acceleration

t = time

Then:

v = 19 m/s - 0.397 m/s² · t

0 = 19 m/s - 0.397 m/s² · t

-19 m/s / -0.397 m/s² = t (acceleration is negative because it is opposite to the direction of the movement)

t = 47.86 s (The truck stoped at 47.86 s after releasing the foot from the accelerator pedal)

With the time and acceleration, we can calculate the traveled distance.

x = 0 m + 19 m/s · 47.86 s - 1/2 · 0.397 m/s² · (47.86s)²

x = 454.66 m (without rounding the acceleration nor the time, the value will be 454.86 m)

Now, we can calculate the work done to stop the truck which will be of the same magnitude as the kinetic energy:

W = 900 N · 454.66 m = 409194 J

(if you do all the calculations without rounding, you will get the same value as we calculated above using the equation of kinetic energy, 409374 J).

3 0
3 years ago
10.20) Perform the following conversions: (a) 0.912 atm to torr,(b) 0.685 bar to kilopascals, (c) 655 mm Hg to atmospheres,(d) 1
masha68 [24]

Answer:

(a) 693.12 torr

(b) 68.5 kilopascals

(c) 0.862 atmosphere

(d) 1.306 atmospheres

(e) 36.74 psi

Explanation:

(a) 0.912 atm = 0.912 atm × 760 torr/1 atm = 693.12 torr

(b) 0.685 bar = 0.685 bar × 100 kPa/1 bar = 68.5 kPa

(c) 655 mmHg = 655 mmHg × 1 atm/760 mmHg = 0.862 atm

(d) 1.323×10^5 Pa = 1.323×10^5 Pa × 1 atm/1.01325×10^5 Pa = 1.306 atm

(e) 2.50 atm = 2.50 atm × 14.696 psi/1 atm = 36.74 psi

5 0
4 years ago
Read 2 more answers
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