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adelina 88 [10]
3 years ago
10

Which of the following is an example of deduction

Physics
1 answer:
Serjik [45]3 years ago
3 0
 a  because of the following rate I do not now I saw it on the other one
You might be interested in
4
Zigmanuir [339]

(a) The object moves with uniform velocity from A to B.

(b) The object moves with constant velocity from B to C.

(c) The object moves with increasing velocity from C to D.

<h3>Velocity of the object from point A to B</h3>

V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s

<h3>Velocity of the object from point B to C</h3>

V(B to C) = (6 - 6)/(11 - 4) = 0 m/s

<h3>Velocity of the object from point C to D</h3>

V(C to D) = (7 - 6)/(12 - 11) = 1 m/s

final velocity = 1 + 1.5 m/s = 2.5 m/s

Thus, we can conclude the following;

The object moves with uniform velocity from A to B.

The object moves with constant velocity from B to C.

The object moves with increasing velocity from C to D.

Learn more about velocity here: brainly.com/question/6504879

#SPJ1

4 0
2 years ago
The formula for speed is v = s/t. <br> a. True<br> b. False
crimeas [40]
If you mean S is the distance then it is true 
Velocity = Distance / time 
3 0
3 years ago
Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the o
blondinia [14]

The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

Q=ne

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C

The potential difference V between the disks separated by a distance d is given by,

V=Ed

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V

The capacitance C of the capacitor is given by,

C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F

The capacitance of a parallel plate capacitor is given by,

C=\frac{\epsilon_0A}{d}

Here, ε₀ is the permittivity of free space  and A is the area of the disks.

Rewrite the expression for A.

C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3

the area A of the disks is given by,

A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }

Here, D is the diameter of the disk.

D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm

The diameter of each disc is found to be 1.32 cm.



7 0
3 years ago
What is the speed of a cheetah if it takes 20 sesonds to run 300 m
natta225 [31]

the answer to your question is 15 :)

8 0
3 years ago
Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st
scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

x(t) = v_0 cos \theta t

where

v_0 is the initial speed

t is the time

\theta=32.0^{\circ} is the angle below the horizontal

We can rewrite this equation as

t=\frac{x(t)}{v_0 cos \theta} (1)

The vertical displacement instead is given by

y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2 (2)

where

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

Substituting (1) into (2),

y(t) = -x(t) tan \theta - \frac{1}{2}gt^2

We know that for t = time of flight, the horizontal displacement is

x(t) =50.8 m

We also know that the vertical displacement is

y(t) = -45 m

Substituting everything into the equation, we can find the time of flight:

\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

x(t) = v_0 cos \theta t

where we have

x = 50.8 m

\theta=32.0^{\circ}

Substituting the time of flight,

t = 1.64 s

We find:

v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s

The initial vertical velocity is instead

u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s

And it changes according to the equation

v_y = u_y -gt

So at t = 1.64 s (when the ball hits the ground),

v_y = -19.3 - (9.8)(1.64)=-35.4 m/s

So the impact speed is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s

While the direction is:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}

8 0
3 years ago
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