Ask question

Ask question

All categories

3 years ago

15

A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al

ong the negative y-axis, all in 1.60 s. Determine the x-component and the y-component of the rabbit's acceleration. (Enter your answers in m/s2. Indicate the direction with the signs of your answers.)

1 answer:

anzhelika [568]3 years ago

8 0

Answer:

aₓ = 0 , ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

vₓ = v₀ₓ = 1.10 m / s

aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

$v_{y}$ = v_{oy} -ay t

ay = (v_{oy} -v_{y}) / t

ay = (0 -10.9) / 1.6

ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

You might be interested in

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

7 0

3 years ago

1. A car is traveling at 36 m/s hits a tree and comes to rest in 0.05 seconds.

QveST [7]

Answer: a. -720m/s^2

b. Yes, airbags will deploy

Explanation:

The formula for acceleration is:

= (Final velocity - Initial velocity)/Time

Final velocity = 0m/s

Initial velocity = 36m/s

Time taken = 0.05s

= (Final velocity - Initial velocity)/Time

= (0 - 36)/0.05

= -36/0.05

= -720m/s^2.

Since it's negative, it shows that there was a deceleration.

2. Yes the airbag will deploy since the acceleration gotten is more than -600 m/s^2.

3 0

3 years ago

Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecule

Colt1911 [192]

Answer:

8563732.58906 Pa

3992793.23326 Pa

5708.00923 J

Explanation:

V = Volume

N = Number of molecules = $3\times 6.023\times 10^{23}$

T = Temperature = 300 K

b = $7\times 10^{-29}\ m^3$

$k_$ = Boltzmann constant = $1.38\times 10^{-23}\ J/K$

P = Pressure

We have the equation

$P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa$

The pressure is 8563732.58906 Pa

For isothermal expansion

$P_1(V_1-Nb)=P_2(V_2-Nb)\\\Rightarrow P_2=\dfrac{P_1(V_1-Nb)}{V_2-Nb}\\\Rightarrow P_2=\dfrac{8563732.58906(0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29})}{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P_2=3992793.23326\ Pa$

The pressure is 3992793.23326 Pa

Work done is given by

$dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J$

The work done is 5708.00923 J

7 0

3 years ago

the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev?

kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s ( this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

E = 4.14 eV

7 0

3 years ago

What is a concern about recovered memories?

Fittoniya [83]

Answer:

D because those are both concerning.

4 0

3 years ago

Read 2 more answers