Most of the symbols used to indicate elements are from an elements name in which language

?/1 Jorge traveled 5 miles north to school. He then traveled 3 miles west to the store. Then he left the store and traveled 5 mi

aleksklad [387]

Answer:

He's 3 miles west of school.

Explanation:

He went 5 miles up and 5 miles down which means that he really didn't go up or down. In between that, he went 3 miles west so if the 5 milers don't count, this puts him at 3 miles west of school.

4 0

3 years ago

A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially

Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0 ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt= 6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>

4 0

3 years ago

A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu

bija089 [108]

Answer:

Explanation:

<u>Impulse-Momentum Theorem</u>

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed $v_1$ and changes it to $v_2$ . The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

$p=mv$

The initial and final momentums are, respectively

$p_1=mv_1,\ p_2=mv_2$

The change of momentum is

$\Delta p=p_2-p_1=m(v_2-v_1)$

It is numerically equal to the Impulse J

$J=\Delta p$

$J=m(v_2-v_1)$

We are given

$m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s$

The impulse the car experiences during that time is

$J=2498(2.6-17.1)=2498(-14.5)$

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

8 0

3 years ago

Jack has two boxes. One is 148g and one is 78g. If jack pushes both boxes with the same amount of force, which will accelerate f

Step2247 [10]

The 78g box, since it has less weight, would accelerate faster. If you had a frictionless surface, and you conducted this experiment, both boxes, without any outside forces, would accelerate at the same rate forever. However, in this problem we must assume the surface is not frictionless. Friction is determined by weight; the more weight, the more friction. Since the 78g box has less weight, it has less friction, making it easier to push with less force.

8 0

4 years ago

From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , suc

marysya [2.9K]

Answer:

q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

ΔV = Δ $V_{C}$

we assume that the source has a voltage of the form

ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

C = q / ΔV

q = C ΔV sin wt

the current in the circuit is

i = dq / dt

i = c ΔV₀ w cos wt

if we use

cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

V_{L} + V_{C} = 0

L di / dt + q / C = 0

the current is

i = dq / dt

they ask us for a solution so that

L d²q / dt² + 1 / C q = 0

d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

di / dt = Aw cos (wt + Ф)

d²i / dt²= - A w² sin (wt + Ф)

- A w² + 1 /LC A = 0

w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

0 = A sin Ф

Ф = 0

q = q₀ sin (wt)

6 0

3 years ago