Ask question

Ask question

All categories

2 years ago

14

Alonzo sprints for 500 meters along a straight track during a race. After crossing the finish line, he to walks back along the t

rack for 50 meters to cool down. What is Alonzo's displacement? 450 meters, straight ahead 500 meters, straight ahead 550 meters, straight ahead

2 answers:

Nata [24]2 years ago

6 0

450 meters, straight ahead is the answer.

allsm [11]2 years ago

3 0

Answer:

450 meters straight ahead

Explanation:

As we know that displacement is the total vector displacement of the object from initial point to final point

As here the Alonzo moves in a straight line during the finish line which is at distance of 500 m

so initial displacement is given as

$d_1 = 500 m$

now for cool down he again moves back 50 meter along the same track in backward direction

so it is given as

$d_2 = -50 m$

so we can say that total displacement will be

$d = 500 - 50 = 450 m$ straight ahead

You might be interested in

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

2 years ago

A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

$\theta = 234.62 radian$

3 0

2 years ago

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

2 years ago

When two objects are in contact, moving together, which of the following statements must be true? Choose all that apply. When tw

Setler [38]

Answer:

The objects must have the same acceleration and the objects must exert the same magnitude force on each other.

Explanation:

The objects must have the same weight: FALSE. This is not needed, any two object can move together in contact no matter their mass.

The objects must have the same acceleration: TRUE. If they have different accelerations, they will separate since the distance each of them travel at a given time will be different.

The objects must have the same net force acting on them: FALSE. This is not needed, since what matters is acceleration, and a=F/m, so if both objects have different net force acting on them, they could have different masses also to compensate and result in the same acceleration.

The objects must exert the same magnitude force on each other: TRUE, this is the 3rd Newton Law, an action must follow the same reaction.

7 0

2 years ago

The current in a long solenoid of radius 6 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of

jonny [76]

Answer:

53.63 μA

Explanation:

radius of solenoid, r = 6 cm

Area of solenoid = 3.14 x 6 x 6 = 113.04 cm^2 = 0.0113 m^2

n = 17 turns / cm = 1700 /m

di / dt = 5 A/s

The magnetic field due to the solenoid is given by

B = μ0 n i

dB / dt = μ0 n di / dt

The rate of change in magnetic flux linked with the solenoid =

Area of coil x dB/dt

= 3.14 x 8 x 8 x 10^-4 x μ0 n di / dt

= 3.14 x 64 x 10^-4 x 4 x 3.14 x 10^-7 x 1700 x 5 = 2.145 x 10^-4

The induced emf is given by the rate of change in magnetic flux linked with the coil.

e = 2.145 x 10^-4 V

i = e / R = 2.145 x 10^-4 / 4 = 5.36 x 10^-5 A = 53.63 μA

6 0

1 year ago