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3 years ago

How far will a brick starting from rest fall freely in 3 seconds

Physics

1 answer:

sertanlavr [38]3 years ago

5 0

Answer:

When object falls freely we can say

its initial speed will be ZERO

and its acceleration due to gravity must be

a = g = 9.8 m/s^2a=g=9.8m/s2

Now we need to find the distance of free fall in 3 s of time

So here we will use kinematics

y = v* t + \frac{1}{2} at^2y=v∗t+21at2

now we will plug in all values in it

y = 0 + \frac{1}{2}* 9.8 * 3^2y=0+21∗9.8∗32

y = 44.1 my=44.1m

So it will fall by total distance of 44.1 m

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A particle that has an 7.3-μC charge moves with a velocity of magnitude 4 × 105 m/s along the +x axis. It experiences no magneti

vovangra [49]

If the particle's velocity were perpendicular to the magnetic field, the magnetic force would be given by:

F = qvB

F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field strength.

Given values:

F = 0.320N

q = 7.3×10⁻⁶C

v = 4×10⁵m/s

Plug in the values and solve for B:

0.320 = (7.3×10⁻⁶)(4×10⁵)B

B = 0.110T

The magnetic force depends on the cross product between the particle's velocity vector v and the magnetic field vector B. If v and B are parallel then the cross product is 0. v points in the +x direction, therefore B must point in either the +x or -x direction.

5 0

3 years ago

Find the voltage gain vO/vS and current gain iO/iX for the circuit for g = 0.0025 S. Then, for vS = 4 V, find the power supplied

Lelechka [254]

Answer:

Incomplete question, no circuit diagram.

Check attachment for further explanation and circuit diagram

Explanation:

We want to find Vo/Vs and Io/Ix and also power delivered to the 2 kΩ resistor

Given that,

g= 0.0025 S (i.e conductance )

Vs= 4 V

R1 = 1 kΩ = 1000 Ω

R2 = 3 kΩ = 3000 Ω

R3 = 10 kΩ = 10000 Ω

R4 = 500Ω = 0.5 kΩ

R5 = 2 kΩ = 2000 Ω

a. At Loop 1: let use voltage divider rule to get Vx

Then, Vx = R2/(R1+R2) • Vs

Vx=3/(1+3) •Vs

Vx=¾ Vs.

The small signal current is given as

Is=g•Vx, since Vx=¾Vs

Then, Is= 0.025×¾Vs

Is=3/1600 Vs. Equation 1

Note: Ressistor R4 and R5 are in series, then the equivalent resistance of R4 and R5 is given as

Req= R4+R5

Req=2+0.5=2.5 kΩ

So, using current divider rule between R3 and the equivalent resistance of R4 and R5.

Therefore, Io= R3/(R3+Req) • Is

Io= R3/(R3+Req) • Is equation 2

Note : using ohms law on resistor R5,

V=iR. , R5=2 kΩ

Vo=IoR5

Vo=2Io

Io=Vo/2. Equation 3

Substitute equation 1 and 3 into 2

Io= R3/(R3+Req) • Is

½Vo = 10/(10+2.5) • 3/1600 Vs

½Vo = 10/12.5 • 3/1600 Vs

½Vo = 3/2000 Vs

Vo/Vs = 3/2000 × 2

Vo/Vs = 1 / 1000

The voltage output gain is

Vo/Vs = 1 / 1000

b. From equation 2

Io= R3/(R3+Req) • Is

Also, applying ohms law to resistor R2,

Vx = Ix• R2, R2=3kΩ

Vx = 3•Ix

Given that, Is= g•Vx

Is=0.0025(3•Ix)

Is= 3/400 Ix

Then, Io= R3/(R3+Req) • Is

Io= 10/(10+2.5) • 3/400 Ix

Io= 10/12.5 • 3/400 Ix

Io= 3/500 Ix

Io/Ix= 3/500

The current gain is

Io/Ix= 3/500

c. Output power

Power is given as

P=I²R

Then, output power at Resistor 5 is

Po = Io²•R5

R5=2000 Ω

From loop 1: using KVL, sum of voltage in a loop is zero

-Vs+1000Ix+3000Ix=0

4000Ix=Vs

Since Vs=4

Then, 4000Ix=4

Ix =4/4000

Ix = 1/1000 A

Since, Io/Ix = 3/500

Then, Io = 3/500 • Ix

Io=3/500 × 1/1000

Io= 6×10^-6 A

Therefore,

Po=Io²•R5. ,R5=2000

Po= (6×10^-6l² × 2000

Po=7.2×10^-8 W

Po=72×10^-9 W

Po=72 nW

The output power at resistor R5 is

72 nW

6 0

3 years ago

The automobile has a weight of 2700 lb and is traveling forward at 4 ft>s when it crashes into the wall. If the impact occurs

alex41 [277]

Answer:

$F_b=153918\, lb.ft.s^{-2}$

Explanation:

Given:

mass, $m=2700 \,lb$

time, $t=0.06\,s$

velocity, $v=4\,ft.s^{-1}$

coefficient of kinetic friction between wheels & pavement, $\mu=0.3$

According to first condition,

$F\times \Delta t=m\times v$

$F\times 0.06= 2700\times 4$

$F=180000\,lb.ft.s^{-2}$

According to second condition,

<u>Magnitude of frictional force (which acts opposite to the direction of motion):</u>

$f=\mu.N$

where N is the normal reaction.

$f=0.3\times 2700\times 32.2$

$f=26082 \,lb.ft.s^{-2}$

Now, the impulsive force on the wall if the brakes were applied during the crash:

$F_b= F-f$

$F_b=180000-26082$

$F_b=153918\, lb.ft.s^{-2}$

3 0

3 years ago

How much heat is needed to raise the temperature of 9g of water by 17oC?

andre [41]

Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7

6 0

3 years ago

1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?

Anettt [7]

The man is holding the suitcase at the same height above the surface of earth. So the gravitation potential energy remains the same.

<span>work done is force * displacement = weight * 0 = 0</span>

7 0

3 years ago