Ask question

Ask question

All categories

4 years ago

8

Two particles A and B start from rest at the origin x = 0 [ft] and move along a straight line such that a = (613) Ift/s) and ag

= (1212.8) [ft/s), where t is in seconds. Determine the distance between them when t3 [s), and the total distance each has traveled in 3 seconds

1 answer:

Amanda [17]4 years ago

3 0

Answer:

Distance between them after 3 s is 2695.5 ft.

Total distance traveled by A in 3 s is 2758.5 ft.

Total distance traveled by B in 3 s is 5454 ft.

Explanation:

For particle A:

u = 0, a = 613 ft/s

Let the distance traveled by particle A in 3 seconds is Sa.

Use second equation of motion

S = u t + 1/2 at ^2

Sa = 0 + 1/2 x 613 x 3 x 3 = 2758.5 ft

For particle B:

u = 0, a = 1212.8 ft/s

Let the distance traveled by particle B in 3 seconds is Sb.

Use second equation of motion

S = u t + 1/2 at ^2

Sb = 0 + 1/2 x 1212 x 3 x 3 = 5454 ft

Thus, the difference in the distance traveled by A and B is 5454 - 2758.5 = 2695.5 ft.

You might be interested in

To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceler

kirza4 [7]

Answer:

a.) 2.66 seconds for 79.2 km/h and 1.82 seconds for 47.8 km/h

b.)-4.13 m/s2 for 79.2 km/h and -3.63 m/s2 for 47.8 km/h

Explanation:

Now for (a)

time = velocity/distance

for velocity = 79.2 km/h and distance 0.0586 km

t = (0.0586/79.2)*3600

t = 2.66 seconds (Please note that multiplication with 3600 is to convert hours into seconds)

for velocity 47.8 km/h and distance 0.0242 km

t = (0.0242/47.8)*3600

t = 1.82 seconds

Now for (b)

2as = vf2-vi2

so,

2a(58.6) = -(79.2*1000/3600)^2 (Please note that multiplication and division with 1000 and 3600 respectively is to convert speed unit from km/h to m/s)

a = -4.13 m/s2 for 58.6 m

and

2a(24.2) = -(47.8*1000/3600)^2

a = -3.63 m/s2 for 24.2 m.

Please note that "-" sign express the deceleration.

5 0

3 years ago

German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (

Ierofanga [76]

Answer:

The uncertainty in the position of the electron is $5.79x10^{-9}m$

Explanation:

The Heisenberg uncertainty principle is defined as:

$\Lambda p\Lambda x$ ≥ $\frac{h}{4 \pi}$ (1)

Where $\Lambda p$ is the uncertainty in momentum, $\Lambda x$ is the uncertainty in position and h is the Planck's constant.

The momentum is defined as:

$p =mv$ (2)

Therefore, equation 2 can be replaced in equation 1

$\Lambda (mv) \Lambda x$ ≥ $\frac{h}{4 \pi}$

Since, the mass of the electron is constant, v will be the one with an associated uncertainty.

$m \Lambda v \Lambda x$ ≥ $\frac{h}{4 \pi}$ (3)

Then, $\Lambda x$ can be isolated from equation 3

$\Lambda x$ ≥ $\frac{h}{m \Lambda v 4 \pi}$ (4)

$\Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}$

But $1J = Kg.m^{2}/s^{2}$

$\Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}$

$\Lambda x = 5.79x10^{-9}m$

Hence, the uncertainty in the position of the electron is $5.79x10^{-9}m$

7 0

3 years ago

List five rules of golf etiquette. Explain the correct stance, grip, orientation, and parts of a swing.

Verizon [17]

- keep you eyes on the ball, after you hit the ball you look up, you have to see the club hit the ball

- your grip it really matters on what your comfortable with but first you put your left hand on the club then put your right hand below your left hand, then put your right pinky on top of your index finger, but you can inter lock your pinky and index finger

- DON'T HIT AT THE BALL HIT THREW THE BALL

- parts of the swing, can be pretty tricky to explain, but I'll try my best to explain it, okay so a golf swing is like a pendulum, you swing right to left. Once you lift the club to the right you still look down at the ball. After swimming to the right your have to hit the ball with your right hand strength, in order the ball to fly.

- the way you do your grip defines if the ball will go straight or not. But also the way you swing towards the ball also defines how far the ball is going to go.

8 0

3 years ago

Read 2 more answers

A pendulum is constructed from a string 0.0530 m long attached to a mass of 0.200 kg. When set into motion, the pendulum complet

Jet001 [13]

To solve this problem we will start from the pendulum period equation. From there we will find the value of gravity. Later it will be possible to apply the linear motion kinematic equations, for which we will use the value of the distance traveled and the gravity in order to find the fall time. The time period of the pendulum is

$T=2\pi\sqrt{\frac{L}{g}}$

Here,

L is the length of the pendulum

g is acceleration due to gravity.

Rearranging to find g,

$g = \frac{4\pi^2 L}{T^2}$

Replacing,

$g = \frac{4\pi^2 (0.0530)}{1.5^2}$

$g = 0.9299m/s^2$

Now using,

$x=v_0 t + \frac{1}{2} gt^2$

Here,

$v_0 =$ Initial velocity (0 because the pendulum is at rest)

t is time taken to cover the distance .

$x = \frac{1}{2}gt^2$

Replacing,

$2.5 = \frac{1}{2} (0.9299) t^2$

Solving for t

$t = 2.3188s$

Therefore it will take to fall through the distance of 2.5m around to 2.31s

5 0

3 years ago

A 3kg box is pulled along the floor with a force of P=20 N at an angle of 30 degrees. If the box is moving at a constant velocit

Deffense [45]

If force is applied at 30 degree with the horizontal

then first we will find the components of force in two directions

$F_x = 20 cos20$

$F_y = 20 sin30$

now we can find normal force by balancing the forces in Y direction

$F_n + Fy = mg$

$F_n + 20 sin30 = 3*10$

$F_n + 10 = 30$

$F_n = 20 N$

now since block is moving with constant speed

so here forces along horizontal plane will be balanced also

$F_x = F_f$

$F_f = 20 cos30$

$F_f = 17.3 N$

Now the weight of the box is defined as

$W = mg$

now plug in all values

$W = 3*10 = 30 N$

7 0

3 years ago