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aleksandr82 [10.1K]
3 years ago
5

True or falseas mass increases the gravitational force decreases​

Physics
2 answers:
blondinia [14]3 years ago
7 0
The answer to this question is true I believe
MrMuchimi3 years ago
7 0
The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them. This means that the force of gravity increases with mass, but decreases with increasing distance between objects.
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A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of
mote1985 [20]

Answer:

<h2>640N/cm^2</h2>

Answer D is correct

Explanation:

pressure =  \frac{force}{area}  \\  =  \frac{320}{0.5}  \\  = 640

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0
3 years ago
What is 34 + (3) × (1.2465) written with the correct number of significant figures?
sergejj [24]
34 + 3.7395 = 37.7395
The answer is 38.
The answer is written to two significant figures because the smallest given  number of significant figure is 2.
6 0
4 years ago
Read 2 more answers
Calculate the work done by a 47 n force pushing a 0.025 kg pencil 0.25 m against a force of 23.
inna [77]
Fnet=fa-fr
=47-23
=24n
W=F∆Xcos°
=24n×0.025cos0
=0.6J
8 0
3 years ago
A car is moving at 76 miles per hour. the kinetic energy of that car is 5 × 105 j. how much energy does the same car have when i
valentina_108 [34]
Its would be 144. cause if u think about it
3 0
3 years ago
The activation energy of a certain reaction is 37.2 kJ/mol . At 20 ∘C, the rate constant is 0.0130 s−1. At what temperature woul
Butoxors [25]

Answer: At 34°c

Explanation:

Using The Arrhenius Equation:

k = Ae − Ea/RT

k represents rate constant

A represents frequency factor and is constant

R represents gas constant which is = 8.31J/K/mol

Ea represents the activation energy

T represents the absolute temperature.

By taking the natural log of both sides,

ln k = ln A- Ea/RT

Reactions at temperatures T1 and T2 can be written as;

ln k1= ln A− Ea/RT1

ln k2= ln A− Ea/RT2

Therefore,

ln(k1/k2) = −Ea/RT1 + Ea/RT2

Since k2=2k1 this becomes:

ln(1/2) = Ea/R*[1/T2 − 1/T1]

Theefore,

-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]

1/T2 - 1/293 = -1.55 x 10^-4

1/T2 = -1.55 x 10^-4 + 34.13x 10^-4

1/T2 = 32.58 x 10^-4

Therefore T2 = 307K

T2 = 307 - 273 = 34 °c

7 0
3 years ago
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