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3 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]3 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.

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3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

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3 years ago

A steel piano string of mass per unit lenght 4×10-⁴kgm-¹ was struck, if the speed of the sound on the string is 300ms-¹. Calcula

attashe74 [19]

Answer:

36 N

Explanation:

Velocity of a standing wave in a stretched string is:

v = √(T/ρ),

where T is the tension and ρ is the mass per unit length.

300 m/s = √(T / 4×10⁻⁴ kg/m)

T = 36 N

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3 years ago

10. Calculate the kinetic energy of a running back that has a mass of 80 kg and

EastWind [94]

Answer:

The answer is

<h2>2560 J</h2>

Explanation:

The kinetic energy of an object given it's mass and velocity can be found by using the formula

$KE = \frac{1}{2} m {v}^{2}$

where

m is the mass

v is the velocity

From the question

m = 80 kg

v = 8 m/s

The kinetic energy is

$KE = \frac{1}{2} \times 80 \times {8}^{2} \\ = 40 \times 64$

We have the final answer as

<h3>2560 J</h3>

Hope this helps you

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3 years ago

A rubber band has a spring constant of 45 N/m. A classmate stretches it a length of 0.2 m before firing it at Mr. C. Ouch. Calcu

nata0808 [166]

Answer:

<h2>0.9 Joules</h2>

Explanation:

Elastic potential energy of a spring= 1/2 × Spring constant × displacement²

following calculations you will get ur answer!!

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