A 7.0-kg bowling ball experiences a net force of 5.0 N.
Whatwill be its acceleration?

Answer:

The mass of the steam is 91.2 g

Mass of the steam=91 grams

Explanation:

Given:

When steam is mixed with the ice then the heat loss by the steam will be gained by the ice so there will no overall heat gain or loss during the mixing

So According to question

Let M be the mass of the steam mixed with ice then we have

[tex]M\times2256\times10^3+M\times4186\times(100-33)=0.499\times222\times10^3+0.499\times4186\times(33-0)\\M\times2.58\times10^6=2.35\times10^5\\M=91.2\ \rm g[/tex]

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:

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.

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:

[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]

Answer:

0.21 m/s/s.

Explanation:

Whenever there is an action force acting on a body, there will be a reaction force.

Here the force with which the astronaut throws the tool is given as 16 N.

Force is measured in newtons and is equal to the rate of change of momentum.

Since the astronaut has a mass, she experience a reaction force. It is given by F = ma, according to Newton's 2nd law.

16 = 75 a

⇒ Acceleration = a = F/m = 16/75 = 0.21 m/s/s

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.

The average speed for the entire trip is 83.6 km/h. The average velocity for the entire trip is zero.

The average speed for the entire trip can be calculated by finding the total distance traveled divided by the total time taken. In this case, the student traveled 39.0 km to school and then returned the same distance back home, resulting in a total distance of 78.0 km. The total time taken for the entire trip is the sum of the time taken to drive to school, the time taken to return home, and the time spent finding the report, which is 23.0 min + 23.0 min + 10.0 min = 56.0 min. To convert the total time to hours, divide by 60: 56.0 min / 60 = 0.933 hours. Therefore, the average speed for the entire trip is 78.0 km / 0.933 hours = 83.6 km/h.

The average velocity for the entire trip can be determined by considering both the magnitude and direction of the displacement. In this case, since the student drove in the same direction to school and back home, the displacement is zero, meaning there was no change in position. Thus, the average velocity for the entire trip is also zero.

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:

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:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

f = Focal length = 35

m = Magnification = 2.5

[tex]m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u[/tex]

Lens equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm[/tex]

[tex]v=-2.5\times 21=-52.5\ cm[/tex]

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

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:

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]

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:

a) The car doesn't hit the barrier because after he sees the lights of a barrier he only travels for 27.95m, enough to miss the barrier. b) The maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s or 75.6km/hr.

Explanation:

a)

In order to solve this problem we must first do a drawing of the situation for us to visualize it better. (See picture attached).

As you may see on the drawing, the car still travels some distance when the driver notices the lights of the barrier. This distance is calculated for a constant velocity V:

x=Vt

the car has a velcity of 60km/h which is equivalent to:

[tex]\frac {60km}{hr} * \frac{1000m}{1km}*\frac{1hr}{3600s}=16.667m/s[/tex]

so on the first 0.80s the car travels a distance of:

x=(16.667m/s)(0.8s)=13.333m

Next, the car breaks, so we can say it moves with a constant acceleration of -9.5m/s, so the distance is found by using the following formula:

[tex]x=\frac {V_{f}^{2}-V_{0}^{2}}{2a}[/tex]

We know the final velocity will happen when the car stops, so the final velocity is zero, leaving us with:

[tex]x=\frac {-V_{0}^{2}}{2a}[/tex]

so we can substitute the provided values to find the distance traveled by the car during this time.

[tex]x=\frac{-(16.667m/s)^{2}}{2(-9.5m/s^{2})}[/tex]

which yields:

x=14.62m

so in total the car traveled:

[tex]x_{tot}=13.333m+14.62m = 27.95m[/tex]

Which is not enough for the car to hit the barrier.

b)

In order to solve this part of the problem, we must combine the two equations we got on the previous part to find a single equation that will represent the total displacement of the car:

[tex]x=vt-\frac{v^{2}}{2a}[/tex]

so we can now substitute the known values so we get:

[tex]40=0.8v-\frac{v^{2}}{2(-9.5)}[/tex]

which simplifies to:

[tex]40=0.8v+0.05263v^{2}[/tex]

this is a quadratic equation so it can be solved by using the quadratic formula, but first we must rewrite it in standard form, so we get:

[tex]0.05263v^{2}+0.8v-40=0[/tex]

now we can use the quadratic formula:

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

So we can substitute the given values:

[tex]v=\frac{-0.8\pm\sqrt{(0.8)^{2}-4(0.05263)(-40)}}{2(0.05263)}[/tex]

which returns two answers:

v=21m/s  and   v=-36.20m/s

We take the positive answer since is the one that represents a moving towards the right side of the drawing.

So the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s

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:

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:

[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]

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:

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:

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:

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:

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]

To find the position, velocity, and acceleration of the snowboarder at t = 12 seconds, evaluate the given displacement equation and its first and second derivatives at t = 12 seconds, then round the final answers to one decimal place.

To determine the position, velocity, and acceleration of the snowboarder at t = 12 seconds, we need to evaluate the given displacement function x(t) = 0.5t³ + t² + 2t and its derivatives at that specific time value.

Position at t = 12 seconds

By substituting t = 12 into the displacement function, we get:
x(12) = 0.5(12)³ + (12)² + 2(12), which can be calculated to give the position x.

Velocity at t = 12 seconds

The velocity v(t) is the first derivative of the displacement function, v(t) = 1.5t² + 2t + 2. Evaluate v(12) to find the velocity at t = 12 seconds.

Acceleration at t = 12 seconds

Acceleration a(t) is the second derivative of the displacement function, a(t) = 3t + 2. Evaluate a(12) to find the acceleration at t = 12 seconds.

Using these steps, you can calculate the exact values and round them to one decimal place as asked in the question.

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) A = [tex]3.90 \AA[/tex]

(b) [tex]A = 4.50 \AA[/tex]

(c) [tex]A = 5.51 \AA[/tex]

(d) [tex]A = 9.02 \AA[/tex]

Solution:

As per the question:

Radius of atom, r = 1.95 [tex]\AA = 1.95\times 10^{- 10} m[/tex]

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = [tex]2\times 1.95 = 3.90 \AA[/tex]

(b) For body centered cubic lattice:

[tex]A = \frac{4}{\sqrt{3}}r[/tex]

[tex]A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA[/tex]

(c) For face centered cubic lattice:

[tex]A = 2{\sqrt{2}}r[/tex]

[tex]A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA[/tex]

(d) For diamond lattice:

[tex]A = 2\times \frac{4}{\sqrt{3}}r[/tex]

[tex]A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA[/tex]

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:

E=[tex]k*\frac{q}{21}*u[/tex]

[tex]u=\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]

[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]

Explanation:

q: particle's charge

k: coulomb constant

E=E*u

r=r*u

r=distancia vectorial entre P y S

r=distancia escalar entre P y S

E: Electric field vector

E: magnitud of magnetic field vector

u: unit vector radial

then:

[tex]E=k*q/r^{2}[/tex]

r=r*u

r=P-S=(-1,-2,-4)m

[tex]r^{2}=(Magnitude(P-S))^2=(-1)^2+(-2)^2+(-4)^2=21[/tex]

[tex]r=\sqrt{21}[/tex]

E=[tex]k*\frac{q}{21}*u[/tex]

u=r/r=[tex]\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]

[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]

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

Final answer:

The volume of the air bubble just before it reaches the surface is 3.32 cm^3.

Explanation:

To find the volume of the air bubble just before it reaches the surface, we can use the ideal gas law equation: PV = nRT. Since the pressure and amount of gas remain constant, we can rewrite the equation as V/T = k, where V is the volume, T is the temperature, and k is a constant.

Using the given temperatures at the bottom and top of the lake (5.9°C and 16.0°C) and the initial volume of the bubble (1.22 cm^3), we can set up the following equation:

(1.22 cm^3) / (5.9°C) = V / (16.0°C).

Solving for V, the volume of the bubble just before it reaches the surface, we get:

V = (1.22 cm^3)(16.0°C) / (5.9°C) = 3.32 cm^3.