Insert two arithmetic means between 13 and 18.5

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straig

kupik [55]

Answer:

Explanation:

V = Deltax/Deltat

V = 15.0 m/s

Displacement:

(a) Vf = Vi + adeltat

Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s

Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m

(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s

Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m

(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s

Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m

(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s = 19.6m/s

Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m

4 0

4 years ago

A bicycle tire is inflated to a gauge pressure of 3.71 atm when the temperature is 16°C. While a man rides the bicycle, the temp

Contact [7]

Answer:

The pressure of the tire will be 4.19 atm

Explanation:

The ideal gas equation is:

p0 * V0 / T0 = p1 * V1 / T1

Since V0 = V1 because the volume of the tire doesn't change

p0 / T0 = p1 / T1

p1 = p0 * T1 / T0

We need absolute temperatures for this equation

16 C = 289 K

45 C = 318 K

Now we replace the values:

p1 = 3.71 * 318 / 289 = 4.19 atm

8 0

3 years ago

A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so

shusha [124]

Answer:

<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>

Explanation:

<u>Accelerated Motion</u>

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

$T_y+N=m.g$

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$