Physics High School

According to Newton's 2nd law of motion, if you hit a baseball with a bat it will
-hurt your hand from the reaction force
-accelerate according to the applied force
-fly in a straight line unless an outside force changes its course
-fall to the ground because of gravitational pull

Answers

Answer 1

Answer:

Newton's second law states that force is directly proportional to the product of mass and acceleration.

so if you hit a baseball with a bat it will accelerate according to the applied force.

Related Questions

Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q₁ and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.500 m. What is net force on 93?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
91
0.500 m-
-5.00 x 10-6 C
92
-5.00 x 10-6 C
93
0.500 m-

Answers

The answer to the question is 1.125  Newton

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.

Electrostatic force between q1 and q3

r will be 0.500 + 0.500 = 1 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²

F₁₃ = 2.25  × 10⁻¹ N

Electrostatic force between q2 and q3

r will be 0.500 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²

F = (225 × 10⁻³) / (25 × 10⁻²)

F₂₃ = 9 × 10⁻¹ N

Total force on q3 will be F₁₃ + F₂₃

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 11.25  × 10⁻¹ ) N

Total force on q3 =  1.125  N

Thus after solving we got the net force on q3 as 1.125  Newton

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What are two models of light? how does each model explain part of the behavior of light?

Answers

The two most successful theories of light were the corpuscular (or particle) theory of Sir Isaac Newton and the wave theory of Christian Huygens. Newton's corpuscular theory stated that light consisted of particles that travelled in straight lines.

hope it helps

A ball is shot from the ground straight up into the air with initial velocity of 40 ft/sec. Assuming that the air resistance can be ignored, how high does it go

Answers

The height to which the ball attained, given the data is 7.58 m

Data obtained from the questionInitial velocity (u) = 40 ft/s = 40 × 0.3048 = 12.192 m/sFinal velocity (v) = 0 m/s (at maximum height) Acceleration due to gravity (g) = 9.8 m/s²Maximum height (h) =?

How to determine the maximum height

v² = u² – 2gh (since the ball is going against gravity)

0² = 12.192² – (2 × 9.8 × h)

0 = 12.192² – 19.6h

Collect like terms

0 – 12.192² = –19.6h

–12.192² = –19.6h

Divide both side by –19.6

h = –12.192² / –19.6

h = 7.58 m

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What is the change of state when magma crystallises to form igneous rock?

Answers

Answer:

liquid

Explanation:

due to high temperatures in the magma chamber all solids liquidify on reaching the surface with colder temperatures it now solidify to form igneous rock

the 12-lead ecg shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv. what is this finding indicative of?

Answers

The 12-lead ECG test that shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, could be a diagnostic for myocardial infarction.

What is 12 - lead ECG test?

A 12-lead electrocardiogram (ECG) is a medical test that is recorded using leads, or nodes, attached to the body.

Electrocardiograms, sometimes referred to as ECGs, capture the electrical activity of the heart and transfer it to graphed paper.

When a 12-lead ECG shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, it could be a diagnostic for myocardial infarction.

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The vertical component of 9N is found to be four-thirds of the horizontal component.
a) Find the horizontal component.
b)Find the magnitude of the vector​

Answers

Answer:

a) Let horizontal component be h

4/3×h= 9

4h = 9x3

h = 27/4

b) Magnitude = √ [9²+ (27/4)²

= 11.25

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