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bazaltina [42]
3 years ago
9

An inductive argument guarantees its conclusion True False

Physics
2 answers:
lorasvet [3.4K]3 years ago
8 0

Answer:

False

Explanation:

inductive arguments have some evidence but do not have full assurance of truth of the conclusion

Soloha48 [4]3 years ago
4 0
I think the answer is true
You might be interested in
What is the acceleration of a 5kg mass pushed by a 10N force?
dem82 [27]

Answer:2m/s^2

Explanation:

mass=5kg

Force=10N

Acceleration=force ➗ mass

Acceleration=10 ➗ 5

Acceleration=2m/s^2

4 0
3 years ago
Find mass when kinetic energy is 1.6 J and velocity is 0.2 m/s
klio [65]

Answer:

4kg

Explanation:

Given parameters:

Kinetic energy  = 1.6J

Velocity  = 0.2m/s

Unknown:

Mass of the body = ?

Solution:

The kinetic energy of a body is the energy due to the motion of the body.

It is mathematically expressed as;

  Kinetic energy = \frac{1}{2}  m v²

m is the mass

v is the velocity

          1.6  =  \frac{1}{2}  x 0.2 x v²  

          1.6  = 0.1v²

         v²   = 16

          v  = 4kg

5 0
3 years ago
2. An archer shoots an arrow at 83.0 m/s at a 62.0 degree angle. If the ground is flat, how much time is the arrow in the air?
UkoKoshka [18]

Answer:

<em>t=14.96 sec</em>

Explanation:

<u>Diagonal Launch </u>

It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.

The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is :

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the initial speed, \theta is the angle, t is the time and g is the acceleration of gravity .

In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus y_o=0, and:

\displaystyle y=v_osin\theta \ t-\frac{gt^2}{2}

The value of y is zero twice: when t=0 (at launching time) and in t=t_f when it goes back to the ground. We need to find that time t_f by making y=0

\displaystyle 0=v_osin\theta\ t_f-\frac{gt_f^2}{2}

Dividing by t_f

\displaystyle v_osin\theta=\frac{gt_f}{2}

Then we find the total flight time as

\displaystyle t_f=\frac{2v_osin\theta}{g}

\displaystyle t_f=\frac{2(83)sin\ 62^o}{9.8}

\displaystyle t_f=14.96\ sec

5 0
3 years ago
determine the magnitude and direction of the resultant for a combination of the vectors A: 10.1m, 60 degrees B: 7.3m, 275 degree
quester [9]

Answer:

5.87 m, 75.5°

Explanation:

Let's say C is the resultant vector.

The x component is the sum of the x components of A and B.

Cx = Ax + Bx

Cx = 10.1 cos 60° + 7.3 cos 275°

Cx = 5.69

The y component is the sum of the y components of A and B.

Cy = Ay + By

Cy = 10.1 sin 60° + 7.3 sin 275°

Cy = 1.47

The magnitude is found with Pythagorean theorem, and the angle is found with trigonometry.

C² = Cx² + Cy²

C² = 5.69² + 1.47²

C = 5.87

θ = atan(Cy / Cx)

θ = atan(5.69 / 1.47)

θ = 75.5°

5 0
3 years ago
Plz help me it is improtant
marishachu [46]
I think it is b cause I don’t think you do that
7 0
3 years ago
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