PHYSICS COLLEGE

Leaving the distance between the 181 kg and the 712 kg masses fixed, at what distance from the 712 kg mass (other than infinitely remote ones) does the 72.6 kg mass experience a net force of zero? Answer in units of m

Answers

Answer 1

Answer:

Explanation:

The force due to gravitation is equal to zero for each of the masses.

M1= 181kg

M2= 712kg

m = 72.6kg

The distance between M1 and M2 is said to be fixed , therefore no value should be given I.e it's a constant.

From the formula for gravitational force we have that

F = GMm/r^2

GmM1/(d-r)^2. = GmM2/r^2

where r is the distance between the 72.6 kg and 712kg

d is the distance between M1 and M2

Solving mathematically

r(√M1+√M2) = d√M2

r = d√M2/√M1 + √M2

d×26.68/ 13.45+26.68

d×26.68/40.13

r = 0.665d

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1. Hoai Nguyen, a Physics 2A student, drop a soccer ball from the roof of the new science building. The ball strikes the ground in 3.30 s later. You may ignore air resistance, so the ball is in free fall. How tall, in meters, is the building? 2. How fast was the ball moving right before hitting the floor?

3. Thu Tran, another Physics 2A student, grabs the ball and kicks it straight up to Hoai Nguyen, who is still up on the building rooftop. Assuming that the ball is kicked at 0.50 m above the ground and it goes on a vertical path, what is the minimum velocity required for the ball to make it to the building rooftop? Ignore air resistance. (Hint: the ball will pass the rooftop level with a higher speed...)

Answers

Answer:

1. 53.415 m

2. 32.373 m/s

3. 30.82 m/s

Explanation:

Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved

1. $s = gt^2/2 = 9.81*3.3^2/2 = 53.415 m$

2. $v = gt = 9.81*3.3 = 32.373 m/s$

3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is

53.415 - 0.5 = 48.415 m

For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed

$v^2 - v_0^2 = 2g\Delta s$

where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, $v_0$ is the initial velocity, $\Delta s = 48.415$ is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity

$0 - v_0^2 = 2*(-9.81)*48.415$

$v_0^2 = 950$

$v_0 = √(950) = 30.82 m/s$

The thickness of a $1 bill is 0.11 mm. If you have a stack of $1 bills 450 m tall, how much money do you have?

Answers

Answer:

4090909

Explanation:

Thickness of one bill = 0.11 mm

Total thickness = 450 m

No of $1 bills = total thickness / thickness of one bill

No of $1 bills = 450 / 0.11 × 10^-3

= 4090909

Answer:

You will have 4.5 million dollar

Explanation:

The thickness of a $1 bill is 0.11 mm

So we have

1 $ = 0.1 mm

0.1 mm = 1 $

0.0001 m = 1 $

1 m = 10000 $

450 m = 450 x 10000 = 4500000 $

So you will have 4.5 million dollar

A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?

Answers

The acceleration of the particle is 38.87 kg.

Net magnetic field

The net magnetic field is calculated as follows;

$B_(net) = √(B_x^2 + B_y^2 + B_z^2) \n\nB_(net) = √(2^2 + 3^2 + 4^2) = 5.385 \ mT$

Magnetic force on the charge

The magnetic force on the charge is calculated as follows;

$F = qvB * sin(\theta)\n\nF = 5* 10^(-9) * 5* 10^3 * 5.385 * 10^(-3) * sin(120)\n\nF = 1.166 * 10^(-7) \ N$

Acceleration of the particle

The acceleration of the particle is calculated as follows;

$a = (F)/(m) \n\na = (1.166 * 10^(-7))/(3 * 10^(-9)) \n\na = 38.87 \ kg$

Learn more about magnetic force here: brainly.com/question/13277365

Explanation:

It is given that,

Charge on the particle, $q=5\ nC=5* 10^(-9)\ C$

Mass of the particle, $m=3\ \mu g=3* 10^(-6)\ g=3* 10^(-9)\ kg$

Magnetic field component, $B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT$

Net magnetic field, $B=√(2^2+3^2+4^2)=5.38\ mT=0.00538\ T$

Speed of the particle, v = 5 km/s = 5000 m/s

Angle between velocity and magnetic field, $\theta=120$

Magnetic force is given by :

$F=qvB\ sin\theta$

$F=5* 10^(-9)* 5000\ m/s* 0.00538* sin(120)$

$F=1.16* 10^(-7)\ N$

Acceleration of the particle is given by, $a=(F)/(m)$

$a=(1.16* 10^(-7)\ N)/(3* 10^(-9)\ kg)$

$a=38.6\ m/s^2$

So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.

Of the four most important pathways by which stress affects health, the first one to occuris usually related to physiology

Answers

Answer: Physiologic response to fear is very similar to that of PTSD and stress. Fear is accompanied by increased heart rate due to the release of adrenaline, sympathetic nervous system is aroused. The release of adrenaline also causes increased sweating, pulse and blood pressure. In line with this, the parasympathetic nervous system experiences reduced activity such as decrease in digestion.

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.50 m from the mirror. The filament is 6.00 mm tall, and the image is to be 37.5 cm tall. Part A: How far in front of the vertex of the mirror should the filament be placed?Part B: to what radius of curvature should you grind the mirror?

Answers

Answer:13.6 cm

Explanation:

Given

v(image distance)=-8.5 m

height of object $(h_1)$ =6 mm

height of image $(h_2)$ =37.5 cm

and magnification of concave mirror is given by $m=(-v)/(u)=(-h_2)/(h_1)$

$m=(-37.5* 10)/(6)=-62.5$

$-62.5=(8.5* 100)/(u)$

u=13.6 cm

so object is at a distance of 13.6 cm from mirror.

for focal length

$(1)/(f)=(1)/(v)+(1)/(u)$

$(1)/(f)=(-1)/(850)+(-1)/(13.6)$

$(1)/(f)=-0.00117-0.0735$

f=-13.4 cm

thus radius of curvature of mirror is R=2f=26.8 cm

Final answer:

The filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror. The radius of curvature for the concave mirror should be approximately 0.85 m.

Explanation:

To determine how far in front of the vertex of the mirror the filament should be placed, we can use the mirror equation:

1/f = 1/do + 1/di

Where f is the focal length of the concave mirror, do is the object distance, and di is the image distance.

With the given information, we have:

do = ?

di = 8.50 m

Using the magnification formula:

magnification = -di/do

By substituting the values we know, we can solve for do:

37.5 cm / 6.00 mm = -8.50 m / do

Solving for do, we find that do ≈ - 0.85 m.

Since the object distance cannot be negative, we conclude that the filament of the headlight lamp should be placed about 0.85 m in front of the vertex of the mirror.

To find the radius of curvature for the concave mirror, we use the mirror formula:

1/f = 1/do + 1/di

With do = -0.85 m and di = 8.50 m, we can rearrange the formula to solve for f:

1/f = 1/-0.85 + 1/8.50

1/f ≈ -1.1765

Solving for f, we find that the focal length is approximately 0.85 m.

Learn more about Concave mirror here:

brainly.com/question/3555871

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What power lens is needed to correct for farsightednesswherethe uncorrrected near point is 75 cm?

Answers

To solve this problem we will apply the concept related to the lens power with which farsightedness can be corrected. Mathematically this value is given by the relationship,

$P = (1)/(f)$