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k0ka [10]
3 years ago
7

What happens to the temperature during an endothermic reaction? ANSWERS; The temperature stays constant. The temperature will de

crease (get colder). The temperature will increase (get warmer)
Physics
1 answer:
EleoNora [17]3 years ago
6 0

Answer:

The temperature will decrease (get colder).

Explanation:

Enthalpy changes are heat changes accompanying physical and chemical changes. The enthalpy change is the difference between the sum of the heat contents of products and the sum of heat contents of reactants.

  • For an endothermic change, heat is absorbed for the reaction.
  • The surrounding becomes colder at the end of the reaction and so is the reaction itself.
  • The right choice is that the temperature will decrease.
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25 equals 1/2×2× V squared
solong [7]

Answer:

v=5

Explanation:

1/2 x 2 is 1

1*5^2 is 25

5^2 is 25

6 0
4 years ago
Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

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

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

5 0
3 years ago
What is moongphaliwala?
Elden [556K]
A person who sells <span>moongphali (groundnuts).</span>
6 0
3 years ago
A constant force is exerted for a short distance on a block of mass that is initially at rest on a horizontal frictionless plane
Irina-Kira [14]

Answer:

v

Explanation:

initial speed is 2v

final speed is 3v

increase is v

7 0
3 years ago
Hanging from a horizontal beam are nine simple pendulums of the following lengths: (a) 0.080, (b) 0.26, (c) 0.49, (d) 0.90, (e)
mote1985 [20]

Answer:

Explanation:

This is the case of forced oscillation . The pendulum having the same or matching time period  or angular frequency with that of angular frequency  of external periodic force , will be in resonance having largest amplitude.

Angular frequency of pendulum having length .9 m

= \sqrt{\frac{g}{l } }

l = .9

angular frequency

= \sqrt{\frac{10}{0.9 } }

= 3.33 rad / s

If we calculate angular frequencies of pendulum of all lengths given , we will find that other lengths do not give angular frequency falling between 2 and 4 radian . So only pendulum having length of .9 m will have vibration of maximum amplitude.

8 0
3 years ago
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