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3 years ago

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3. Question

2 answers:

aalyn [17]3 years ago

5 0

2. Newton's laws of motion

miskamm [114]3 years ago

5 0

Hello!

Answer: Newton's Laws of Motion.

Explanation:

↓↓↓↓↓↓↓↓↓↓↓↓

Isaac Newton he performed many experiments involving light (optics). Then he published works on mathematics, history, calculus, and theology. And he published his most significant book, Philosophiae Naturalis Principia Mathematica, in 1687. Newton's first law of motion states that an object at rest stays at rest, and an object in motion stays in motion, unless acted on by an unbalanced force. This is also known as the law of inertia. Inertia is the natural tendency of objects to resist change in motion. Newton's second law of motion states that the total net force acting an object is equal to mass times acceleration. Newton's third law of motion states that for every action, there is an equal and opposite reaction. It considered action and reaction forces.

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A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.

Kay [80]

Answer:

a.6.5025 J

b.6.5025 J

Explanation:

We are given that

Mass of pellet,m=0.27 g= $0.27\times 10^{-3} kg$

1 kg=1000 g

Spring constant,k=1800 N/m

x=8.5 cm= $8.5\times 10^{-2} m$

1m=100 cm

a.Potential energy stored in the compressed spring is given by

P.E= $\frac{1}{2}kx^2$

$P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2$

$P.E=6.5025 J$

b.By using law of conservation of energy

P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J

6 0

3 years ago

In the drawing, water flows from a wide section of a pipe to a narrow section. In which part of the pipe is the volume flow rate

guapka [62]

Answer:

Volume flow rate is the same in both sections of the pipe

Explanation:

7 0

2 years ago

Redox reactions involve the gain or loss of _____.A. neutronsB. phosphateC. electronsD. protons

vovangra [49]

Redox reactions involve the gain or loss of C. electrons

7 0

3 years ago

The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s) + 3I2 (s) → 2Al3+ (aq) + 6I− (aq) The emf

Triss [41]

Answer: +2.10V

Explanation:

$2Al(s)+3I_2(s)\rightarrow 2Al^{3+}(aq)+6I^-(aq)$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log K$

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log [Al^{3+}]^2\times [I^-]^6$

where,

$E^o_{cell}$ = standard emf for the cell = +2.20 V

n = number of electrons in oxidation-reduction reaction = 6

$E_{cell}$ = emf of the cell = ?

$[Al^{3+}]$ = concentration = $5.0\times 10^{-3}M$

$[I}^{-}]$ = concentration = $0.10M$

Now put all the given values in the above equation, we get:

$E_{cell}=+2.20-\frac{0.059}{6}\log [5.0\times 10^{-3}]^2\times [0.10]^6$

$E_{cell}=2.10V$

The standard emf for the cell using the overall cell reaction below is +2.10 V

5 0

3 years ago

a charge of 7.2x10^-5C is placed in an Electric field with a strength of 4.8x10^5N/C. if the Electric potential energy of the ch

Alecsey [184]

Answer:

$d= 2.2 m$

Explanation:

The data given in the question is

Charge : q = $7.2 \times 10^{-5} C$

Electric field Strength : E = $4.8 \times 10^{5} N/C$

Electrical potential Energy : U = $75J$

Let, the distance be "d"

So, the formula for Electrical potential energy is

$U= q \times E \times d$

Simplified formula for distance become

$d=\frac{U}{q \times E}$

Now , insert the value

$d=\frac{75 J}{7.2 \times 10^{-5}C \times 4.8 \times 10^5 N/C}$

or,

$d=2.17 m$

Rounding off

$d= 2.2 m$

3 0

3 years ago