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enyata [817]
3 years ago
8

Indicate whether the statement is true or false. The point located at the geometric center of an object is called the object's c

enter of gravity.
a. True
b. False
Physics
1 answer:
zmey [24]3 years ago
4 0
This statement is true. The point located at the geometric center of an object is indeed called the object's center of gravity. Center of gravity is the average of the weight of a resultant of the parallel forces, including all the particles that is passing through its body.
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125 cm of gas are collected at 15 °C and
aleksklad [387]
The new volume will be ≈26mL
rounded to two significant figures.
Explanation:
This question involves the combined gas law. The equation to use is:
When working with gas laws, the temperature is always in Kelvins. To get Kelvins from a Celsius temperature, add 273.15 to the Celsius temperature.
6 0
2 years ago
A ball is thrown against a wall and bounces back toward the thrower with the same speed as it had before hitting the wall. Does
lorasvet [3.4K]

Answer:

A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.

Explanation:

4 0
3 years ago
I need to show my work as well but on the computer so, please show work for how you got the answers. Thank you!
balandron [24]

Answer:

1) 1H_2 + 1Cl_2 => 2HCl

2) 2Al + 6HCl => 2AlCl_3 + 3H_2

3) 1Ca_2Br_2 + 2NaCO_3 => 2CaCO_3 + 2NaBr

4) 3NaOH + 1FeCl_3 => 3NaCl + 1Fe(OH)_3

Explanation:

4 0
3 years ago
A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece
seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0
2 years ago
When an athlete holds a barbell overhead, the reaction force is the weight of the barbell on his hand. how does this force vary
Sonja [21]
The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
3 0
3 years ago
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