If you break quartz to learn if it splits smoothly in a certain direction, what physical property are you testing?

A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A

natta225 [31]

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

5 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago

A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C

GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0

3 years ago

Radio signals travel at a rate of 3 × 108 meters per second. how many seconds would it take for a radio signal to travel from a

nlexa [21]

<span>3.2x10^-2 seconds (0.032 seconds)
This is a simple matter of division. I also suspect it's an exercise in scientific notation, so here is how you divide in scientific notation:

9.6 x 10^6 m / 3x10^8 m/s

First, divide the significands like you would normally.
9.6 / 3 = 3.2

And subtract the exponent. So
6 - 8 = -2

So the answer is 3.2 x 10^-2
And since the significand is less than 10 and at least 1, we don't need to normalize it.

So it takes 3.2x10^-2 seconds for the radio signal to reach the satellite.</span>

6 0

3 years ago

The radius of the wheel is 40cm and the radius of the

snow_tiger [21]

Answer:

M.A = load/ effort

1200N/400N

= 3

velocity ratio= radius of wheel/radius of Axle

40cm/10cm

=4

efficiency= 3/4*100

75%

4 0

3 years ago