Ask question

Ask question

All categories

3 years ago

9

At t= 0, a particle moving in the xy plane with constant acceleration has a velocity of Vi= (3.00i -2.00j) m/s and is at the ori

gin. At t= 3.00 s, the particle's velocity is V = (9.00i + 7.00j) m/s . Find (a) the acceleration of the particle and (b) its coordinates at any time t.

1 answer:

ivanzaharov [21]3 years ago

4 0

Answer:

(a) a = (2i + 4.5j) m/s^2

(b) r = ro + vot + (1/2)at^2

Explanation:

(a) The acceleration of the particle is given by:

$\vec{a}=\frac{\vec{v}-\vec{v_o}}{t}\\\\$

vo: initial velocity = (3.00i -2.00j) m/s

v: final velocity = (9.00i + 7.00j) m/s

t = 3s

by replacing the values of the vectors and time you obtain:

$\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2$

(b) The position vector is given by:

$\vec{r}=\vec{r_o}+\vec{v_o}t+\frac{1}{2}\vec{a}t^2$

where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2

You might be interested in

A plastic tube allows a flow of 15.9 cm3 /s of water through it. how long will it take to fill a 237 cm3 bottle with water? answ

xxMikexx [17]

You are given a fixed rate of 15.9 cm³/s. You are also given with the amount of volume in 237 cm³. Through the approach of dimensional analysis, you can manipulate through operations such that the end result of the units must be in seconds. The solution is as follows:

237 cm³ * (1 s/15.9 cm³) = 14.9 seconds

7 0

3 years ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago

A spherical gas-storage tank with an inside diameter of 9 m is being constructed to store gas under an internal pressure of 1.50

lutik1710 [3]

Answer: 33 mm

Explanation:

Given

Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m

Internal pressure of gas, P(i) = 1.5 MPa

Yield strength of steel, P(y) = 340 MPa

Factor of safety = 0.3

Allowable stress = 340 * 0.3 = 102 MPa

σ = pr / 2t, where

σ = allowable stress

p = internal pressure

r = radius of the tank

t = minimum wall thickness

t = pr / 2σ

t = 1.5*10^6 * 4.5 / 2 * 102*10^6

t = 0.033 m

t = 33 mm

The minimum thickness of the wall required is therefore, 33 mm

6 0

3 years ago

Scientific way of thinking

noname [10]

Answer:

huh,? can you explain the question more please

7 0

3 years ago

Read 2 more answers

How much force is needed to accelerate a 1000-kg car at a rate of 3 m/s squared

Burka [1]

Answer:

3000 newton force is required

Explanation:

F = ma

F= 1000 kgs x 3 m/s^ 2

F=3000(kgs x m/s^2)

F=3000 newton

8 0

3 years ago