PHYSICS COLLEGE

Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. show answer No Attempt Approximately what is the force due to the Bernoulli effect on a roof having an area of 205 m2? Typical air density in Boulder is 1.14 kg/m3 , and the corresponding atmospheric pressure is 8.89 × 104 N/m2 . (Bernoulli’s principle assumes a laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)

Answers

Answer 1

Answer:

Answer:

The force exerted on the roof is $F =2.37*10^(5)N$

Explanation:

From the question we are told that

The speed of the wind is $v = 45.0 m/s$

The area of the roof is $A = 205 m^2$

The air density of Boulder is $\rho = 1.14 kg / m^3$

The atmospheric pressure is $P_(atm) = 8.89 * 10^(4) N/ m^2$

For a laminar flow the Bernoulli’s principle is mathematically represented as

$P_1 + (1)/(2) \rho v_a ^2 + \rho g h_a = P_2 + (1)/(2) \rho v_b ^2 + \rho h_b$

Where $v_1$ is the speed of air in the building

$v_b$ is the speed of air outside the building

$P_1 \ and \ P_2$ are the pressure of inside and outside the house

$h_a \ and \ h_b$ are the height above and below the roof

Now for $h_a = h_b$

The above equation becomes

$P_1 + (1)/(2) \rho v_a ^2 = P_2 + (1)/(2) \rho v_b ^2$

$P_1 - P_2 = (1)/(2) \rho (v_b^2 - v_a^2)$

Since pressure is mathematically represented as

$P = (F)/(A )$

The above equation can be written as

$F = (1)/(2) \rho ( v_b^2 - v_a ^2 ) A$

The initial velocity is 0

Substituting value

$F = (1)/(2) (1.14) [(45^2 - 0^2 ) ](205)$

$F =2.37*10^(5)N$

Answer:

0.46Ω

Explanation:

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $(V^(2) )/(R)$

=> R = $(V^(2) )/(P)$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $(V^(2) )/(P)$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $(12.0^(2) )/(4)$

R₁ = $(144)/(4)$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $(V^(2) )/(P)$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $(12.0^(2) )/(4)$

R₂ = $(144)/(4)$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$(1)/(R_(X) )$ = $(1)/(R_1)$ + $(1)/(R_2)$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$(1)/(R_X)$ = $(1)/(36)$ + $(1)/(36)$

$(1)/(R_X)$ = $(2)/(36)$

Rₓ = $(36)/(2)$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $(11.7)/(18)$

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $(0.3)/(0.65)$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Answer:

$R_i_n_t=0.45 \Omega$

Explanation:

Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

$R_i_n_t=((V_N_L)/(V_F_L) -1)R_L$

Where:

$V_F_L=Load\hspace{3}voltage=11.7V\nV_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\nR_L=Load\hspace{3}resistance$

As you can see, we don't know the exactly value of the $R_L$ . However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

$(12)/(11.7) =(4)/(P)$

Solving for $P$ :

$P=(11.7*4)/(12) =3.9W$

Now, the electric power is given by:

$P=(V^2)/(R_b)$

Where:

$R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb$

So:

$R_b=(V^2)/(P) =(11.7^2)/(3.9) =35.1\Omega$

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

$(1)/(R_L) =(1)/(R_b) +(1)/(R_b) =(2)/(R_b) \n\n R_L= (R_b)/(2) =(35.1)/(2)=17.55\Omega$

Finally, now we have all the data, let's replace it into the internal resistance equation:

$R_i_n_t=((12)/(11.7) -1)17.55=0.45\Omega$

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

Answers

The minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Given data:

The weight of ladder is, W = 100 N.

The length of ladder is, L = 8.0 m.

The coefficient of static friction between ladder and floor is, $\mu =0.40$ .

Apply the Newton' law in vertical direction to obtain the value of Normal Force (P) as,

$N = mg$

And force along the horizontal direction is,

$F= \mu * N\n\nF = \mu * mg$

Now, taking the torque along the either end of ladder as,

$-mgcos \theta * (L)/(2)+Fsin \theta * L =0\n\nmgcos \theta * (L)/(2) = Fsin \theta * L$

Solving as,

$mgcos \theta * (L)/(2) = (\mu mg) * sin \theta * L\n\ncos \theta * (1)/(2) = (\mu) * sin \theta\n\ntan \theta = (1)/(2 * 0.40 )\n\n\theta = tan^(-1)(1/0.80)\n\n\theta = 51.34^(\circ)$

Thus, we can conclude that the minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Learn more about the frictional force here:

brainly.com/question/4230804

Answer:

The minimum angle is 51.34°

Explanation:

Given that,

Weight of ladder = 100 N

Length = 8.0 m

Coefficient of static friction = 0.40

We need to calculate the normal force

Using Newtons law in vertical direction

$F_(y)=n-mg$

$N-mg=0$

$N=mg$

We need to calculate the normal force

Using Newtons law in horizontal direction

$F_(s)=f_(s)-P$

$f_(s)-P=0$

$f_(s)=P$

$P=\mu mg$

We need to calculate the minimum angle

Using torque about the point A then

$-mg\cos\theta* AB+P\sin\theta* AC=0$

Put the value into the formula

$mg\cos\theta*((L)/(2))=\mu mg\sin\theta* L$

$\cos\theta*(1)/(2)=\mu\sin\theta$

$(1)/(2)=\mu*\tan\theta$

$\theta=\tan^(-1)((1)/(2*0.40))$

$\theta=51.34^(\circ)$

Hence, The minimum angle is 51.34°

On a map each inch represents 25 miles what is the length of a highway if it is 6 inches long on a map

Answers

Answer:

150 hope this helps

Explanation:

Answer:

150

Explanation:

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

Answers

The magnitude of the speed is 83.0325 m\s, the direction is 62.7 degrees, and the fraction of kinetic energy lost is 0.895.

What is collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

$\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}$

The momentum (P₂) after a collision will be

$\rm P_2 = (m_1 + m_2) u \ cos\ \theta \ \hat{x} \ + (m_1 + m_2) u \ sin \ \theta \ \hat{y}$

Applying momentum conservation, we have

$\rm (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\ \theta \ \n\n$ ...1

$\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \ \theta$ ...2

From equations 1 and 2, we have

$\rm \theta = tan \ ^(-1) ( (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2))/( (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2))\n\n\n\theta = tan \ ^(-1) (5.12*239*cos21.3+3.05*282*cos15.4)/(5.12*239*sin21.3-3.05*282*sin15.4)\n\n\n\theta = 62.7^o$

From equation 1, we have

$\rm u = ((m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) )/( (m_1 + m_2) \ cos\ \theta ) \n\n\nu = (5.12*239*cos21.3 - 3.05*282*cos15.4)/((5.12+3.05)cos62.2)\n\n\nu = 83.0325 m/s$

Then the change in kinetic energy, we have

$\rm \Delta KE = (1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2-(1)/(2)(m_1+m_2)u^2\n\n\n\Delta KE = (1)/(2) * 5.12 * 239^2 + (1)/(2)*3.05*282^2 - (1)/(2)(5.12+3.05)*83.032^2\n\n\n\Delta KE = 239.34 \ J$

The fraction of kinetic energy lost will be

$\rm Energy \ lost = (239.34)/(267.5) = 0.895$

More about the collision link is given below.

brainly.com/question/13876829

Answer:

Detailed solution is given below

A disk of mass 5 kg and radius 1m is rotating about its center. A lump of clay of mass 3kg is dropped onto the disk at a radius of 0.5m , sticking to the disk. If the system is rotating with an angular velocity of 11 rad/s, what is the final angular momentum of the disk h the clay lump?wit? ( Idisk = MR^2/2)

Answers

Answer:

$27.5 kgm^2/s$

Explanation:

We can solve for the final angular velocity of the system using the law of momentum conservation

$I_1\omega_1 = I_2\omega_2 = M_2$

Where $I_1 = MR^2/2 = 5*1^2/2 = 2.5 kgm^2$ is the moments of inertia of the disk before. $I_2 = I_1 + mr^2 = 2.5 + 3*0.5^2 = 2.5 + 0.75 = 3.25 kgm^2$ is the moments of inertia of the disk after (if we treat the clay as a point particle). $\omega_1 = 11rad/s$ is the angular speed before.

$2.5*11 = M_2$

$M_2 = 27.5 kgm^2/s$

So the final momentum of the system is 27.5 kgm2/s

Answer:

The final angular momentum is 35.75 kg.m²/s

Explanation:

Given;

mass of disk, M = 5 kg

radius of disk, R = 1 m

mass of clay, M = 3 kg

radius of clay, R = 0.5 m

final angular momentum, $\omega _f$ = 11 rad/s

Final angular momentum angular momentum of the disk that the clay lumped with;

$P = I_f\omega_f$

where;

$I_f$ is the final moment of inertia

$I_f = I_(disk) + I _(sand)\n\nI_f = (M_DR^2)/(2) + M_SR^2\n\nI_f = (5*1^2)/(2)+ 3*0.5^2\n\nI_f = 2.5 + 0.75=3.25 \ kg.m^2$

Final angular momentum of the disk;

= $I_f \omega_f$

= 3.25 x 11 = 35.75 kg.m²/s

Therefore, the final angular momentum is 35.75 kg.m²/s

The gravitational force law, deduced by Newton in the 1660's, is remarkably similar to Coulomb's law. Recall that the universal law of gravitation states that the magnitude of the gravitational force between two masses M1 and M2 separated by a distance R is given by the following equation:________. F = G (M1 x M2) / R2
G = 6.67 x 10-11 Nm2/kg2
a. Calculate the value of the gravitational force between an electron (mass = 9.11 x 10-31 kg) and a proton (mass is 1836 times greater than the mass of an electron) if the two particles are separated by 3.602 nanometers. (1 nanometer or 1 nm = 1 x 10-9 m)
F= ______ N
b. The force created in the above question is:
1. repulsive
2. attractive