The area under the v-t graph represent the ___

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0

3 years ago

1. What are the 2 main categories of nonmetals?

Anettt [7]

the 2 main categories of nonmetals are REACTIVE NONMETALS an NOBLE GAS.

hope it helps

7 0

3 years ago

A thin rod of length 0.75 m and mass 0.42 kg is suspended

MrRissso [65]

Answer:

a) K = 0.63 J, b) h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

w² = $\frac{m g d}{I}$

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

I = ⅓ m L²

we substitute

w = $\sqrt{\frac{mgL}{2} \ \frac{1}{\frac{1}{3} mL^2} }$

w = $\sqrt{\frac{3}{2} \ \frac{g}{L} }$

w = $\sqrt{ \frac{3}{2} \ \frac{9.8}{0.75} }$

w = 4.427 rad / s

an oscillatory system is described by the expression

θ = θ₀ cos (wt + Φ)

the angular velocity is

w = dθ /dt

w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

w = 4.0 rad / s

the kinetic energy is

K = ½ I w²

K = ½ (⅓ m L²) w²

K = 1/6 m L² w²

K = 1/6 0.42 0.75² 4.0²

K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

Em₀ = K = ½ I w²

final point. Highest point

Em_f = U = m g h

energy is conserved

Em₀ = Em_f

½ I w² = m g h

½ (⅓ m L²) w² = m g h

h = 1/6 L² w² / g

h = 1/6 0.75² 4.0² / 9.8

h = 0.153 m

5 0

2 years ago

Time Intervals in Ice Ages

stepladder [879]

It started off with 68% less than it did at the peak, and later created a void and melted the remainder of the ice at about 92%

8 0

3 years ago

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

1 year ago

The area under the v-t graph represent the ____ of an object