What does an excess of electrons produce?

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 disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg per

GREYUIT [131]

Answer:

Explanation:

To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .

I₁ = 1/2 mr²

= .5 x 175 x 2.13²

= 396.97 kgm²

I₂ = m r²

= 55.4 x 2.13²

= 251.34 mgm²

ω₁ = .651 rev /s

= .651 x 2π rad /s

ω₂ = tangential velocity of man / radius of disc

= 3.51 / 2.13

= 1.65 rad/s

I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω

396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω

ω = 3.14 rad /s

kinetic energy = 1/2 I ω²

= 3196 J

8 0

4 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago

The angle between incident ray and reflected ray is 80 degrees. What is the value of angle of incidence?

saul85 [17]

Answer:

The angle of incident ray is 40°.

Explanation:

Given that the angle of incident and reflected ray are the same. In this question, we had given that both angles added up will gives you 80° so you have to divide it by 2 :

incident + reflected = 80°

Let incident = reflected = θ

θ + θ = 80°

2θ = 80°

θ = 80° ÷ 2

= 40°

8 0

3 years ago

A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away

Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d) Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it. Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball. Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.

Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

x = x(0) + V_i,x*t_c

D = 0 + V_i,x*t_c

V_o,x + v_r = D/t_c

V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

V_o = sqrt ( V_o,x^2 + V_o,y^2)

V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

- The angle with which it should be thrown can be evaluated by trigonometric relation:

tan(Q) = ( V_o,y / V_o,x )

tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

6 0

3 years ago