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OlgaM077 [116]
3 years ago
7

Which of the following is an example of exothermic reaction?

Physics
1 answer:
Katarina [22]3 years ago
6 0

Explanation:

Exothermic reaction are those in which heat releases during a reaction

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Cars with hybrid and electric motors what?
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What are the periodic variations in Earth's rotation and orbit around the sun that alter the way solar radiation is distributed
Dahasolnce [82]

Answer:

1. The precession of the equinoxes.

2. Changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun.

3. Variations in the eccentricity

Explanation:

These variations listed above;  the precession of the equinoxes (refers, changes in the timing of the seasons of summer and winter), this occurs on  a roughly about 26,000-year interval; changes in the tilt angle of Earth’s rotational axis relative to the plane of Earth’s orbit around the Sun, this occurs roughly in a 41,000-year interval; and changes in the eccentricity (that is a departure from a perfect circle) of Earth’s orbit around the Sun, occurring on a roughly 100,000-year timescale. which influences the mean annual solar radiation at the top of Earth’s atmosphere.

5 0
3 years ago
A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin
Mariana [72]

Answer:

The velocity of block = 0.188 \frac{m}{s}

Explanation:

Mass m = 5.6 kg

k = 1040 \frac{N}{m}

\mu = 0.26

x_{1} = 0.035 m  , v_{1} = 0

x_{2} = 0.02 m

From work energy theorem

K_{1} + U_{1} + W_{other} = K_{2} + U_{2}  --------- (1)

Kinetic energy

K = \frac{1}{2} k x^{2}  ------- (1)

Potential energy

U = \frac{1}{2} k x^{2} ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

R_{N} = mg

R_{N}  = 5.6 × 9.81 = 54.9 N

Friction  force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

W_{f} = - 21.9 × 0.015

W_{f} = - 0.329 J

Now kinetic energy

At point 1

K_{1} = \frac{1}{2} m v^{2} _{1}

K_{1} = 0

U_{1} = \frac{1}{2} k x^{2}

U_{1} = \frac{1}{2}  (1040) 0.035^{2}

U_{1} = 0.637 J

At point 2

K_{2} = \frac{1}{2} (5.6) v^{2} _{2}

K_{2} = 2.8 v_{2} ^{2}

Potential energy

U_{2} = \frac{1}{2}  k x_2^{2}

U_{2} = \frac{1}{2}  (1040) 0.02^{2}

U_{2} = 0.208 J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 v_{2} ^{2} + 0.208

2.8 v_{2} ^{2} = 0.1

v_{2} = 0.188 \frac{m}{s}

This is the velocity of block.

6 0
4 years ago
ASAP PLZZ FINALS TOMR GIVE U ALOT OF POINTSS
Annette [7]

Answer:

B

Explanation:

Let m be mass of the object and v be speed of object b.

Kinetic Energy of B = 1/2 mv^2

Kinetic Energy of A = 1/2 m(2v)^2

= 2 mv^2

= 4 (1/2 mv^2)

= 4 × Kinetic Energy of B

Hence Object A has four times the kinetic energy of object B (<em>A</em><em>n</em><em>s</em><em> </em><em>B</em><em>)</em>

7 0
4 years ago
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