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3 years ago

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An arrow strikes a target moving at 75 m/s and embeds itself 15 cm into the target. If the arrow stopped with constant accelerat

ion, how long (in seconds) after striking the target did it take for the arrow to come to rest?

1 answer:

zhenek [66]3 years ago

5 0

Answer:

Answer:

4 ms

Explanation:

initial velocity, u = 75 m/s

final velocity, v = 0

distance, s = 15 cm = 0.15 m

Let the acceleration is a and the time taken is t.

Use third equation of motion

v² = u² + 2 a s

0 = 75 x 75 - 2 a x 0.15

a = - 18750 m/s^2

Use first equation of motion

v = u + at

0 = 75 - 18750 x t

t = 4 x 10^-3 s

t = 4 ms

thus, the time taken is 4 ms.

Explanation:

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What does the slope of a secant line represent?

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Answer:

A secant line is a straight line joining two points on a function. It is also equivalent to the average rate of change, or simply the slope between two points. The average rate of change of a function between two points and the slope between two points are the same thing.

Explanation:

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3 years ago

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.959 g, q = 5.84 µC is locat

NemiM [27]

Answer:

$8.66\times 10^{-6}\ C$ or $8.66\ \mu C$ .

Explanation:

<u>Given:</u>

Charge on the particle at origin = Q.
Mass of the moving charged particle, $\rm m = 0.959\ g = 0.959\times 10^{-3}\ kg.$
Charge on the moving charged particle, $\rm q = 5.84\ \mu C = 5.84\times 10^{-6}\ C.$
Distance of the moving charged particle from first at t = 0 time, $\rm r=20.7\ cm = 0.207\ m.$
Speed of the moving particle, $\rm v = 47.9\ m/s.$

For the moving particle to circular motion, the electrostatic force between the two must be balanced by the centripetal force on the moving particle.

The electrostatic force on the moving particle due to the charge Q at origin is given by Coulomb's law as:

$\rm F_e = \dfrac{kqQ}{r^2}.$

where, $\rm k$ is the Coulomb's constant having value $\rm 9\times 10^9\ Nm^2/C^2.$

The centripetal force on the moving particle due to particle at origin is given as:

$\rm F_c = \dfrac{mv^2}{r}.$

For the two forces to be balanced,

$\rm F_e = F_c\\\dfrac{kqQ}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow Q = \dfrac{mv^2}{r}\times \dfrac{r^2}{kq}\\=\dfrac{mv^2r}{kq}\\=\dfrac{(0.959\times 10^{-3})\times (47.9)^2\times (0.207)}{(9\times 10^9)\times (5.84\times 10^{-6})}\\=8.66\times 10^{-6}\ C\\=8.66\ \mu C.$

6 0

3 years ago

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

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3 years ago

What is the unit of measurement of velocity

melomori [17]

Usually the unit of measurement of velocity is meters per second or m/s

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3 years ago

Read 2 more answers

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while

Vinvika [58]

Answer:

Explanation:

Given

Time taken to reach ground is $t=2.7\ s$

Malda initial velocity $u=95\ cm/s$

Let h be the height of Cliff

using $h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

$h=0+\frac{1}{2}(9.8)(2.7)^2$

$h=35.72\ m$

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

$R=u\times t$

$R=0.95\times 2.7=2.43\ m$

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3 years ago