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DIA [1.3K]
3 years ago
8

A student performs a reaction twice. In the second trial, he increases the temperature of the reaction and notes that the reacti

on happens more quickly. The student concludes that this means the reaction was endothermic. What is wrong with the student's reasoning?
Higher temperature increases reaction rates for both endothermic and exothermic reactions.
An endothermic reaction would not speed up at higher temperatures.
Temperature does not affect the rate of a reaction.
The higher temperature in the second trial actually lowered the activation energy of the reaction.
Physics
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer: higher temperature increases reaction rates for both endothermic and exothermic reactions

Explanation: i took the test

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Veseljchak [2.6K]
A :-) for this question , we should apply
F = ma
Given - F = 12 N
a = 0.20 m/s^2
Solution -
F = ma
12 = m x 0.20
m = 12 by 0.20
m = 60 kg

.:. The mass is 60 kg.
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When will electric charges flow?
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C. when the circuit is closed
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How could you weaken the force of gravity between cars and the Earth?<br>**I WILL MARK BRAINLEST**​
scoundrel [369]

Answer:

The answer you have selected is correct.

Explanation:

Increase radius, force of gravity decreases

6 0
3 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
2 years ago
Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, p
Studentka2010 [4]

Answer:

True

Explanation:

This is a representation of Gauss law.

Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.

8 0
3 years ago
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