Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the building, as shown. If Keisha's line of sight makes a 30° angle with the side of the building, what is Keisha's height above the ground? Assume Monique is 1.5 m tall. A. 14 m B. 15 m C. 16 m D. 17 m

Answer:

[tex]Density = 538 \frac{g}{m3} [/tex]

Explanation:

To get the density you need the vapor pressure for the moisture, to get this first you need to find in tables (found in internet, books, apps) the saturation vapor pressure for water at 50°C:

[tex]P_{sat} = 12350 Pa  @ 50°C[/tex]

Now, a relative humidity of 6,5% means that the actual vapor pressure is 6,5% that of the saturated air so:

[tex]P_{vapor} = 12350 Pa * 0,065 = 802,75 Pa = 0,792 atm [/tex]

According to the ideal gases formula density can be calculated as:

[tex]d = \frac{P*M}{R*T}[/tex]

Where:

Ideal gases constant [tex]R = 0,082 \frac{atm L}{mol k}[/tex]

Pressure P = 0,792 atm

Temperature T = 50°C = 323 K

molar mass M = 18 g/mol

[tex]d = 0,538 \frac{g}{L} = 538 \frac{g}{m3} [/tex]

Answer:

The flux of the electric field  is 677.6 Nm²/C

Explanation:

Given that,

Area = 18 cm²

Charge = 6.0 nC

We need to calculate the flux of the electric field

Using Gauss's law

[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]

Where, q = charge

[tex]\epsilon_{0}[/tex] =permittivity of free space

Put the value into the formula

[tex]\phi=\dfrac{6.0\times10^{-9}}{8.854\times10^{-12}}[/tex]

[tex]\phi=677.6\ Nm^2/C[/tex]

Hence, The flux of the electric field  is 677.6 Nm²/C.

Answer:

Explanation:

A) we know that volume is given as V

[tex]V  =\frac{\pi}{4} D^2 h[/tex]

where D = 1.5 in , h = 2.0 in

so [tex]V = \frac{\pi}{4} 1.5^2\times 2 = 3.53in^3[/tex]

[tex]Area =\frac{\pi}{4} D^2 = \frac{\pi}{4} \times 1.5^2 = 1.76 in^4[/tex]

yield strenth is given as[tex] \sigma_y = \frac{force}{area} = \frac{140,000}{1.76}[/tex]

[tex]\sigma_y = 79.224 ksi[/tex]

b)

elastic strain[tex] \epsilon = \frac{\sima_y}{E} = \frac{79.224}{30\times 10^3} = 0.00264[/tex]

strain offsets  = 0.00264 + 0.002 = 0.00464     [where 0.002 is offset given]

[tex]\frac{\delta}{h} = 0.00464[/tex]

[tex]\frac{h_i -h_o}{h_o} = 0.00464[/tex]

[tex]h_i = 2\times(1-0.00464) = 1.99 inch[/tex]

area [tex]A = \frac{volume}{height} = \frac{3.534}{1.9907} = 1.775 in^2[/tex]

True strain[tex] \sigma = \frac{force}{area} = \frac{140,000}{1.775 in^2} = 78,862 psi[/tex]

At P= 260,000 lb ,[tex] A = \frac{3.534}{1.6} = 2.209 inc^2[/tex]

true stress [tex]\sigma  = \frac{260,000}{2.209} = 117,714 psi[/tex]

true strain [tex]\epsilon = ln\frac{2}{1.6} = 0.223[/tex]

flow curve is given as \sigma = k\epsilon^n

[tex]\sigma_1 = 78,862 psi[/tex]

[tex]\epsilon_1 = 0.00464[/tex]

[tex]\sigma_2 = 117,714 psi[/tex]  

[tex]\epsilon_2 = 0.223[/tex]

so flow curve is

[tex]78,868 = K 0.00464^n[/tex] .........1

[tex]117,714 = K 0.223^n[/tex]   .........2

Solving 1 and 2

we get

n = 0.103

and K =137,389 psi

Strength coffecient = K = 137.389ksi

strain hardening exponent = n = 0.103

Answer:

The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]

Explanation:

According to Einstein's photoelectric equation we have

[tex]E_{incident}\geq \phi [/tex]

where

[tex]E_{incident}[/tex] is the energy of the incident light

[tex]\phi [/tex] is the work function of the metal

The incident energy of the light with wavelength [tex]\lambda [/tex] is given by

[tex]E_{incident}=h\cdot \frac{c}{\lambda}[/tex]

Thus the photoelectric equation reduces to

[tex]h\cdot \frac{c}{\lambda}\geq \phi\\\\h\cdot c\geq \lambda \times \phi\\\\\therefore \lambda\leq \frac{h\cdot c}{\phi}[/tex]

Thus applying values we get

[tex]\lambda\leq \frac{6.62\times 10^{-34}\times 3\times 10^{8}}{3.10\times 1.602\times 10^{-19}}\\\\\therefore \lambda\leq 0.4\times 10^{-6}m[/tex]

Hence The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]

The electron can make angles of 118° and 62° with the magnetic field.

An electron moving at a speed of 4.00 × 10³ m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40 × 10-16 N. What angle does the velocity of the electron make with the magnetic field?

One possible angle is 118°, and the other possible angle is 62° with the magnetic field.

The correct answer is [tex]\(\boxed{89.9999999999999\°}\).[/tex]

To determine the angle between the electron's velocity and the magnetic field, we can use the formula for the magnitude of the magnetic force on a moving charge, which is given by the Lorentz force law:

[tex]\[ F = qvB \sin(\theta) \][/tex]

where:

- [tex]\( F \)[/tex] is the magnitude of the magnetic force,

- [tex]\( q \)[/tex] is the charge of the electron,

- [tex]\( v \)[/tex] is the speed of the electron,

- [tex]\( B \)[/tex] is the magnitude of the magnetic field, and

- [tex]\( \theta \)[/tex] is the angle between the electron's velocity and the magnetic field.

The charge of an electron is [tex]\( 1.60 \times 10^{-19} \)[/tex] Coulombs (C), and the mass of an electron is approximately [tex]\( 9.11 \times 10^{-31} \)[/tex] kilograms (kg). Using Newton's second law, [tex]\( F = ma \)[/tex], where [tex]\( m \)[/tex] is the mass of the electron and [tex]\( a \)[/tex] is its acceleration, we can equate the magnetic force to the mass times acceleration:

[tex]\[ qvB \sin(\theta) = ma \][/tex]

Given that the electron experiences only a magnetic force, we can solve for [tex]\( \sin(\theta) \)[/tex]:

[tex]\[ \sin(\theta) = \frac{ma}{qvB} \][/tex]

Plugging in the given values:

[tex]\[ \sin(\theta) = \frac{(9.11 \times 10^{-31} \text{ kg})(2.30 \times 10^{14} \text{ m/s}^2)}{(1.60 \times 10^{-19} \text{ C})(7.69 \times 10^{6} \text{ m/s})(9.21 \times 10^{-4} \text{ T})} \][/tex]

[tex]\[ \sin(\theta) = \frac{(9.11 \times 2.30)}{(1.60 \times 7.69 \times 9.21)} \times 10^{-31 + 14 - (-19) - 6 - (-4)} \][/tex]

[tex]\[ \sin(\theta) = \frac{(20.953)}{(119.9424)} \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]

[tex]\[ \sin(\theta) = 0.1747 \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]

[tex]\[ \sin(\theta) = 0.1747 \times 10^{-10} \][/tex]

[tex]\[ \sin(\theta) = 1.747 \times 10^{-11} \][/tex]

Now, we can find the angle [tex]\( \theta \)[/tex] by taking the inverse sine (arcsin) of [tex]\( \sin(\theta) \)[/tex]:

[tex]\[ \theta = \arcsin(1.747 \times 10^{-11}) \][/tex]

Since the value of [tex]\( \sin(\theta) \)[/tex] is extremely small, the angle [tex]\( \theta \)[/tex] will be very close to 0 degrees. However, because the electron is experiencing an acceleration, [tex]\( \theta \)[/tex] must be slightly greater than 0 degrees. Using a calculator, we find:

[tex]\[ \theta \approx \boxed{89.9999999999999\°} \][/tex]

This result indicates that the electron's velocity is nearly parallel to the magnetic field, with an angle that is almost 90 degrees but infinitesimally less.

Answer:

Δf=73Hz

Explanation:

From the question we know that:

C = 1482 m/s

Vs = 0 m/s

Vr = -4.95 m/s (it's negative because it is in the opposite direction to the waves)

f0 = 22000 Hz

Applying the formula for the doppler effect:

[tex]f=(\frac{C-Vr}{C-Vs} )*fo[/tex]

f = 22073 Hz.   So the difference is only 73Hz

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

[tex]X(van)=5.65t+154[/tex]

[tex]X(driver)=34.4t+\frac{(-2)t^{2} }{2}[/tex]

or by rearanging the drivers equation.

[tex]X(driver)=34.4t+t^{2}[/tex]

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

[tex]X(van)=X(driver)[/tex]

[tex]5.65t+154=34.4t-t^{2}[/tex]

[tex]0=t^{2} -(34.4-5.65)t+154[/tex][tex]0=t^{2} -28.75t+154[/tex]

To solve this equation we use the following formulas

[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]

Where a=1; b=-28.75; c=154

So we get:

[tex]t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s[/tex][tex]t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s[/tex]

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

[tex]V(driver)=V_{0} +at[/tex]}

[tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}[/tex][tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}[/tex]

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer  A).

Best of luck

Answer:

[tex]q=1.18*10^{-6}C}[/tex]

Explanation:

The potential V due to a charge q,  at a distance r, is:

[tex]V=k\frac{q}{r}[/tex]

k=8.99×109 N·m^2/C^2      :Coulomb constant

We solve to find q:

[tex]q=\frac{V*r}{k}=\frac{6.25*10^{2}*17}{8.99*10^{9}}=1.18*10^{-6}C[/tex]

Explanation:

Given that,

Charge 1, [tex]q_1=-5.45\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=4.39\times 10^{-6}\ C[/tex]

Distance between charges, r = 0.0209 m

1. The electric force is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}[/tex]

F = -492.95 N

2. Distance between two identical charges, [tex]r=0.0209\ m[/tex]

Electric force is given by :

[tex]F=\dfrac{kq_3^2}{r^2}[/tex]

[tex]q_3=\sqrt{\dfrac{Fr^2}{k}}[/tex]

[tex]q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}[/tex]

[tex]q_3=4.89\times 10^{-6}\ C[/tex]

Hence, this is the required solution.

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

[tex]F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}[/tex].

Now, making the substitutions

[tex]d \ = \ 2.00 * 10 ^{-15} \ m[/tex],

[tex]q_1 = q_2 = 1.60 * 10 ^ {-19} \ C[/tex],

[tex]\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}[/tex],

we can find:

[tex]F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}[/tex].

[tex]F \ = 57.536 N[/tex].

Not so big for everyday life, but enormous for subatomic particles.

The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton

[tex]\texttt{ }[/tex]

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

[tex]\texttt{ }[/tex]

There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

Let's tackle the problem now !

[tex]\texttt{ }[/tex]

Given:

distance between protons = d = 2 × 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

Unknown:

electric force = F = ?

Solution:

[tex]F = k \frac{q_1 q_2}{(d)^2}[/tex]

[tex]F = k \frac{q^2}{(d)^2}[/tex]

[tex]F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(2 \times 10^{-15})^2}[/tex]

[tex]F = 57.6 \texttt{ Newton}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Static Electricity

Answer:

The cars pass next to one another after 25.28 s.

Explanation:

When the cars pass next to one another, the position of both cars is the same relative to the center of the system of reference (marker 0 in this case). Then:

Position of car 1 = position of car 2

The position of an accelerating object moving in a straight line is given by this equation:

x = x0 +v0 t +1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

If the position of car 1 = position of car 2 then:

0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²

Note that the acceleration of car 2 has to be positive because the car is slowing down and, in consequence, the acceleration has to be opposite to the velocity. The velocity is negative because the direction of car 2 is towards the origin of our system of reference. Let´s continue:

0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²

1200 m - 50.0 m/s * t + 0.10 m/s² * t² = 0

Solving the quadratic equation:

t = 25.28 s

t = 474. 72 s We discard this value because, if we replace it in the equation of the position of car 2, we will get a position of 20762 m, which is impossible because the position of car 2 can´t be greater than 1200 m.

Then, the cars pass next to one another after 25.28 s  

Answer:1.1 eV

Explanation:

Given

Energy(E)=2.9 eV

Work function of the target=1.8 eV

We know that

[tex]Energy(E)=W_0+KE_{max}[/tex]

Where [tex]W_0=work\ function[/tex]

[tex]2.9=1.8+KE_{max}[/tex]

[tex]KE_{max}=2.9-1.8=1.1 eV[/tex]

Answer:

(a) Your right-hand thumb must be pointed in the direction of the south pole.

Explanation:

A solenoid is a coil of insulated copper wire which behaves like a magnet when an electric current passes through it and it loses its magnetic behaviour just after the current flow stops.

Since it behaves a magnet, the magnetic poles must be determined. In order to determine the magnetic pole, we put the coil in our right hand, curl our four fingers in the direction of current flow and then the direction in which our thumb points is the north pole. This is known as the Right-hand rule.

From the above discussion, option (a) does not keep pace with the statement of right-hand rule.

Hence, option (a) is not true.

Answer:

l= 4 mi   : width of the park

w= 1 mi  : length of the park

Explanation:

Formula to find the area of ​​the rectangle:

A= w*l       Formula(1)

Where,

A is the area of the  rectangle in mi²

w is the  width of the rectangle in mi

l is the  width of the rectangle in mi

Known data

A =  4 mi²

l = (w+3)mi    Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l  

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values ​​of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w  = 1 mi  :width of the park

We replace w  = 1 mi  in the equation (1) to calculate the length of the park:

l=  (w+3) mi

l= ( 1+3) mi

l= 4 mi

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

[tex]F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}[/tex]

Where:

G= 6.674×10^−11 N · (m/kg)2

k =  8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

[tex]\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}[/tex]   --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is  5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

[tex]\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}[/tex]

Therefore

[tex]\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}[/tex]

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

Answer:

(a) -3.173 m/s^2

(b) 3.94 s

(c) 2.47 s

Explanation:

initial velocity, u = 25 m/s

final velocity, v = 0

distance, s = 98.5 m

(a) Let a be the acceleration of the car

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

0 = 625 + 2 x a x 98.5

a = -3.173 m/s^2

(b) v = 12.5 m/s

u = 25 m/s

a = - 3.173 m/s^2

Let the time is t.

Use first equation of motion

v = u + a t

12.5 = 25 - 3.173 t

t = 3.94 s

(c) s = 98.5 / 2 = 49.25 m

u = 25 m/s

a = - 3.173 m/s^2

Let the time be t.

Let v be the velocity at this distance.

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=25^{2}+2\times {-3.173}\ times 49.25tex]

v = 17.17 m/s

Use first equation of motion

v = u + at

17.17 = 25 - 3.173 x t

t = 2.47 s

Answer:

So car will go 50 km in 15 minutes

Explanation:

We have given speed of the car = 200 km/hour

Time t = 15 minutes

We know that 1 hour = 60 minute

So 15 minute = [tex]\frac{15}{60}=0.25hour[/tex]

We have to find the distance

We know that distance = speed ×time = 200×0.25=50 km

So car will go 50 km in 15 minutes

So option (d) will be the correct option

Answer:19.3 km,[tex]\theta =44.40^{\circ}[/tex] south of east

Explanation:

Given

Hiker treks [tex]30 ^{\circ}[/tex] south of east at a speed of 15 m/s for 30 min and then turns due to west and hikes at speed of 8 m/s for another 20 min

Let position vector of Hiker at the end of 30 min

[tex]r=27000cos30\hat{i}-27000sin30\hat{j}[/tex]

after he turns west so new position vector of hiker is

[tex]r'=27000cos30\hat{i}-27000sin30\hat{j}-9600\hat{i}[/tex]

[tex]r'=13782.68\hat{i}-13500\hat{j}[/tex]

Therefore Displacement is given by |r'|

[tex]|r'|=\sqrt{13782.68^2+13500^2}[/tex]

[tex]|r'|=19,292.8035 m\ or 19.3 km[/tex]

for direction

[tex]tan\theta =\frac{13500}{13782.68}[/tex]

[tex]\theta =44.40^{\circ}[/tex] south of east

Answer:

Thus the distance traveled horizontally is 116.28 m

Solution:

As per the question:

Distance covered in 26 steps = 20 m

Thus

Distance covered in one step, step size = [tex]\frac{20}{26}[/tex]

Total steps taken = 152

Now, the distance covered in 152 steps = [tex]one step size\times 152[/tex]

The distance covered in 152 steps, D = [tex]\frac{20}{26}\times 152 = 116.92 m[/tex]

The above distances are on a slope of [tex]6^{\circ}[/tex] above horizontal.

Thus the horizontal component of this distance is given by:

[tex]d_{H} = Dcos6^{\circ}[/tex]

where

[tex]d_{H}[/tex] = horizontal component of distance, D

[tex]d_{H} = 116.92cos6^{\circ} = 116.28 m[/tex]

Answer:

The answer is [tex] 3.994 \times 10^{13}\ electrons[/tex]

Explanation:

The amount of negative charge that must be removed is

[tex]\Delta = Final\ charge - initial\ Charge = 2.6 - (-3.8) = 6.4 \ \mu C = 6.4 \times 10^{-6}\ C[/tex]

and the charge of one electron is

[tex]1 e = 1.60217662\times 10^{-19} \ C[/tex]

So the amount of electrons we need to remove is

[tex]x = \frac{6.4 \times 10^{-6}}{1.60217662\times 10^{-19}} \approx 3.994 \times 10^{13}\ electrons[/tex]

Answer:

B .Limited types of data Insufficient data

Explanation:

To analyse and making decision from big data we have following thing

- Sufficient data

- data analyst

- Accurate data

- data Privacy

- data storage

So, limited data or data that cannot be copied is not and obstacle in data handling. hence option B Limited types of data Insufficient data is correct.

Answer:

ball hit the ground from her feet is 1.83 m far away

Explanation:

given data

speed = 5.3 m/s

angle = 12°

height = 1 m

to find out

how far from her feet ball hit ground

solution

we consider here x is horizontal component and y is vertical component

so in vertical

velocity will be = v sin12

vertical speed u = 5.3 sin 12 = 1.1 m/s downward

and

in horizontal , velocity we know v = 5.3 m/s

so from motion of equation

s = ut + 0.5×a×t²

s is distance t is time a is 9.8

put all value

1 = 1.1 ( t) + 0.5×9.8×t²

solve it we get t

t = 0.353 s

and

horizontal distance is = vcos12 × t

so horizontal distance = 5.3×cos12 × ( 0.353)

horizontal distance = 1.83 m

so ball hit the ground from her feet is 1.83 m far away

Answer:

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

[tex]v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s[/tex]

Velocity at the end of its upward acceleration is 138.6 m/s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m[/tex]

Maximum height the rocket reaches is 762.3 m

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s[/tex]

The velocity with which the rocket crashes to the Earth is 122.3 m/s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s[/tex]

Total time from launch to crash is 12.47+11 = 24.47 seconds

Final answer:

The rocket's velocity at the end of its upward acceleration is 138.6 m/s. The maximum height it reaches is 1353.2 m. The rocket crashes to Earth with a velocity of -1.0 m/s.

Explanation:

(a)  To find the velocity at the end of the rocket's upward acceleration, we can use the formula:

v = u + at

Where:
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration = 12.6 m/s2
t = time = 11.0 s

Substituting the values, we get:

v = 0 + 12.6 * 11.0 = 138.6 m/s

Therefore, the velocity at the end of its upward acceleration is 138.6 m/s.

(b)  To find the maximum height the rocket reaches, we can use the second equation of motion:

s = ut + 0.5at2

Where:
s = distance
u = initial velocity
a = acceleration
t = time

In this case, we need to consider the time taken for both the upward acceleration and free fall.

For the upward acceleration:

u = 0 m/s (since the rocket starts from rest)
a = 12.6 m/s2
t = 11.0 s

Substituting the values, we get:

s1 = 0 * 11.0 + 0.5 * 12.6 * (11.0)2 = 388.5 m

For the free fall:

u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = ?

To find the time for free fall, we can use the equation:

u = at

Substituting the values, we get:

138.6 = -9.8t

Solving for t, we get:

t = -14.1 s

However, time cannot be negative in this case. So, we take the absolute value of t:

t = 14.1 s

Substituting the values in the equation for free fall distance, we get:

s2 = 138.6 * 14.1 + 0.5 * (-9.8) * (14.1)2 = 964.7 m

The maximum height reached by the rocket is s1 + s2 = 388.5 m + 964.7 m = 1353.2 m.

(c)  To find the velocity at which the rocket crashes to Earth, we again consider the free fall phase. Using the equation:

v = u + at

Where:
u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = 14.1 s

Substituting the values, we get:

v = 138.6 - 9.8 * 14.1 = -1.0 m/s

The velocity at which the rocket crashes to Earth is -1.0 m/s. The negative sign indicates that the velocity is directed downward.

(d)  The total time from launch to crash is the sum of the time for upward acceleration (11.0 s) and the absolute value of the time for free fall (14.1 s). Therefore, the total time is 11.0 s + 14.1 s = 25.1 s.

Answer:

Magnitude of electric field at mid way is [tex]5.07\times 10^{7} N/C[/tex]

Solution:

As per the question:

Q = [tex]- 5.6\mu m = - 5.6\times 10^{- 6} C[/tex]

Q' = [tex]5.8\mu m = 5.8\times 10^{- 6} C[/tex]

Separation distance, d = 9.0 cm = 0.09 m

The distance between charges at mid-way, O is [tex]\farc{d}{2} = 0.045 m[/tex]

Now, the electric at point O due to charge Q is:

[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]

[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E =  (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E = - 2.49\times 10^{7} N/C[/tex]

Here, negative sign is indicative of the direction of electric field which is towards the point O

Now, the electric at point O due to charge Q' is:

[tex]E' = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]

[tex]E' =  (9\times 10^{9})\frac{5.8\times 10^{- 6}}{(0.045^{2}}[/tex]

[tex]E' = 2.58\times 10^{7} N/C[/tex]

Refer to Fig 1.

Since, both the fields are in the same direction:

[tex]E_{net} = E + E' = 2.49\times 10^{7} + 2.58\times 10^{7} = 5.07\times 10^{7} N/C[/tex]

The magnitude of the electric field at a point midway between two charges can be calculated using the formula E = k * (Q1-Q2) / r^2. Plugging in the values -5.6μC, +5.8μC, and 9.0cm, the magnitude of the electric field is -2.03 x 10^4 N/C.

The magnitude of the electric field at a point midway between two charges can be calculated using the formula:

E = k * (Q1-Q2) / r^2

where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

In this case, the charges are -5.6μC and +5.8μC, and the distance is 9.0cm (0.09m).

Plugging these values into the formula:

E = (9 x 10^9 Nm^2/C^2) * ((-5.6μC) - (+5.8μC)) / (0.09m)^2

Simplifying the equation:

E = -2.03 x 10^4 N/C

Answer:

F = 1080 N

Explanation:

given,

mass of the hammer = 1.8 kg

velocity of the hammer = 6 m/s

distance into board = 30 mm = 0.03 m

to calculate average resistance force = F

kinetic energy of the hammer is equal to the work done by the hammer

[tex]\dfrac{1}{2}mv^2 = Force\times displacement[/tex]

[tex]\dfrac{1}{2}mv^2 = F\times d[/tex]

[tex]\dfrac{1}{2}\times 1.8\times 6^2 = F\times 0.03[/tex]

F = 1080 N

hence, the average resistance force is equal to F = 1080 N

Answer:

Explanation:

At the topmost position,  the car does not have zero velocity but it has velocity of v so that

v² /r = g or centripetal acceleration should be equal to g ( 9.8 )

Considering that,  the car must fall from a height of 2r + h where

mgh = 1/2 mv²

= 1/2 m gr

So h = r/2

Hence the ball must fall from a height of

2r + r /2

= 2.5 r . So that it can provide velocity of v  at the top where

v² / r = g .

Answer:

[tex]f_p = 5000 N[/tex]

Explanation:

GIVEN DATA:

pressure ratio = length ratio

force = 5 N

scale = 1:10

velocity = 1.5 m/s

[tex]B_p = 20 m[/tex]

[tex]h_p = 2m[/tex]

As pressure ratio = length ratio so we have

[tex]\frac{p_m}{p_p} =\frac{l_m}{l_p} =\frac{1}{10}[/tex]

[tex]\frac{f_m *A_m}{f_p *A_p} ==\frac{1}{10}[/tex]

[tex]\frac{f_m}{f_p} * \frac{A_p}{A_m} = \frac{1}{10}[/tex]

[tex] \frac{f_m}{f_p} * \frac{b_p*h_p}{b_m*h_m} =\frac{1}{10}[/tex]

[tex]5 * \frac{1}{\frac{1}{10}} *\frac{1}{\frac{1}{10}} = \frac{f_p}{10}[/tex]

solving F_p

[tex]f_p = 5000 N[/tex]

To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. If both charges are positive, the net force on a third charge placed on the x-axis will never be zero.

a) To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. The net force will be zero when the two forces are equal in magnitude but opposite in direction. Using Coulomb's law, we can calculate the force between the first charge and the third charge at an unknown position x3. We can then set that force equal to the force between the second charge and the third charge at the same position x3, and solve for x3.

b) If both charges are positive, the net force on a third charge placed on the x-axis will never be zero. Positive charges repel each other, so the forces will always be in the same direction.

Answer: 5 m/s^2

Explanation: In order to solve this question we have to use the kinematic equation given by:

Vf= Vo+a*t  where V0 is zero.

we know that it takes Vf( 30 m/s) in 6 seconds

so

a=(30 m/s)/6 s= 5 m/s^2

Answer:

[tex]v(t)=2.7*e^{0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7-1.3k\\\\[/tex]

For t=2.7s and k=0.3 m/s:

[tex]v(2.7s)=1.80m/s[/tex]

For s=6m and k=0.3 m/s:

[tex]v(6m)=6.69m/s\\\\[/tex]

Explanation:

Definition of acceleration:

[tex]a=\frac{dv}{dt} =0.5kv[/tex]

we integrate in order to find v(t):

[tex]\frac{dv}{v} =-0.5kdt[/tex]

[tex]\int\limits^v_0 { \frac{dv}{v}} \, =-0.5k\int\limits^t_0 {dt} \,[/tex]

[tex]ln(v)=-0.5kt+C\\\\ v=A*e^{-0.5kt}[/tex]        A=constant

Definition of velocity:

[tex]v=\frac{ds}{dt} =A*e^{-0.5kt}[/tex]

We integrate:

[tex]v=\frac{ds}{dt} \\s=- (2/k)*A*e^{-0.5kt}+B[/tex]       B=constant

But:

[tex]v=A*e^{-0.5kt}[/tex]⇒[tex]s= -(2/k)*v+B[/tex]

[tex]v(s)=-(\frac{k}{2} )(s-B)=D-\frac{k}{2}*s[/tex]          D=other constant

Initial conditions:t = 0 are s0 = 2.6 m and v0 = 2.7 m/sec:

[tex]v(t)=A*e^{-0.5kt}\\ 2.7=Ae^{-0.5k*0}\\ 2.7=A\\[/tex]

[tex]v(s)=D-\frac{k}{2}*s\\2.7=D-\frac{k}{2}*2.6\\D=2.7+1.3k[/tex]

So:

[tex]v(t)=2.7*e^{-0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7+1.3k\\\\[/tex]

For t=2.7s and k=0.3 m/s:

[tex]v(2.7s)=2.7*e^{-0.5*0.3*2.7}=1.80m/s[/tex]

For s=6m and k=0.3 m/s:

[tex]v(6m)=\frac{0.3}{2}*6+2.7+1.3*0.3=6.69m/s\\\\[/tex]