The electric field of a charge is defined by the force on: An electron A probe charge A proton. A source charge.

Answer:

6.4 J

Explanation:

m = mass of the bullet = 10 g = 0.010 kg

v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s

v'  = final velocity of the bullet after collision = 1 km/s = 1000 m/s

M = mass of the block = 5 kg

V = initial velocity of block before collision = 0 m/s

V'  = final velocity of the block after collision = ?

Using conservation of momentum

mv + MV = mv' + MV'

(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'

V' = 1.6 m/s

Kinetic energy of the block after the collision is given as

KE = (0.5) M V'²

KE = (0.5) (5) (1.6)²

KE = 6.4 J

Final answer:

Using the conservation of momentum, and given that the bullet exits the wooden block with a lower speed, we calculate the block's final speed to be 7.5 m/s after the bullet exits.

Explanation:

To solve this problem, we need to apply the law of conservation of momentum. The law states that if no external forces act on a system, the total momentum of the system remains constant, even though individual objects within the system might change velocities. In this case, the system is made up of the bullet and the wooden block. Before the collision, the block is at rest, so its initial momentum is 0. The bullet’s initial momentum is its mass times its velocity (momentum = mass × velocity). After the collision, the bullet exits the block with a lower speed, indicating that some of its momentum has been transferred to the block.

Step 1: Calculate the initial momentum of the bullet:
Initial momentum of the bullet = mass of bullet × initial velocity of bullet
= 0.05 kg × 500 m/s
= 25 kg·m/s

Step 2: Calculate the final momentum of the bullet:
Final momentum of the bullet = mass of bullet × final velocity of bullet
= 0.05 kg × 200 m/s
= 10 kg·m/s

Step 3: Calculate the change in momentum of the bullet:
Change in momentum = Initial momentum - Final momentum
Change in momentum = 25 kg·m/s - 10 kg·m/s
= 15 kg·m/s

Step 4: Calculate the final momentum of the block:
Since the block was initially at rest, its final momentum is equal to the change in momentum of the bullet (due to conservation of momentum). Therefore, the final momentum of the block is 15 kg·m/s.

Step 5: Calculate the final velocity of the block:
Using the final momentum and the mass of the block, we can find the final velocity:
Final velocity of the block = Final momentum of block / mass of block
= 15 kg·m/s / 2 kg
= 7.5 m/s

Thus, the speed of the block after the bullet has come out the other side is 7.5 m/s.

Answer:

ΔT / Δx = 771 K/m

ΔT = 771 x 0.0475 = 36.62 k

Explanation:

P = 31700 W, A = 0.819 m^2, Δx = 0.0475 m, K = 50.2 W /m k

Use the formula of conduction of heat

H / t = K A x ΔT / Δx

So, ΔT / Δx = P / K A

ΔT / Δx = 31700 / (50.2 x 0.819)

ΔT / Δx = 771 K/m

Now

ΔT = 771 x 0.0475 = 36.62 k

Answer:

[tex]\lambda = 260.9nm[/tex]

Explanation:

As we know that work function is defined as the minimum energy of photons required to produce the photo electric effect

so here we have

[tex]\phi = \frac{hc}[\lambda}[/tex]

now we know that

[tex]h = 6.63\times 10^{-34}[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

[tex]\lambda[/tex] = wavelength of photons

now from above formula

[tex]7.59 \times 10^{-19} = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{\lambda}[/tex]

[tex]\lambda = \frac{(6.36 \times 10^{-34})(3\times 10^8)}{7.59 \times 10^{-19}}[/tex]

[tex]\lambda = 260.9nm[/tex]

In this exercise we have to use the knowledge of finance to calculate the equilibrium value and the quantity, so we have:

Organizing the information given in the statement we have:

So equating the two given equations we have:

[tex]-x^2 - 3x + 80 = 7x+5\\x^2 +3x + 7x + 5 - 80 = 0\\x^2 + 10x - 75 = 0\\x^2- 5x + 15x -75 = 0\\x(x-5) + 15(x-5) = 0[/tex]

So we can see that the roots will be x = 5 and x = -15 since the quantity cannot be in negative therefore, the equilibrium quantity will be = 5 So replace that value at:

[tex]p = -(5)^2-3(5) + 80 = 40\\p = 7(5) + 5 = 40[/tex]

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Taking the drag force of air into account, it takes longer for the ball to travel to the top of its motion compared to the time it takes to fall back down.

When a ball is thrown straight up, the drag force of air acts against its motion. As the ball ascends, the drag force opposes its upward velocity, causing a reduction in the net force acting on the ball. This results in a slower upward acceleration and a longer time to reach the top of its motion.

On the descent, the gravitational force aids the motion, overcoming the drag force. The downward velocity increases, and the time taken to fall back down is less than the time taken to ascend.

In the absence of air resistance, the ascent and descent times would be equal. However, with air resistance considered, the ascent time is prolonged, impacting the overall motion of the ball.

Answer:

(4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])

Explanation:

[tex]\underset{v_{i}}{\rightarrow}[/tex] = initial velocity of the baseball before collision = 43 [tex]\hat{i}[/tex] m/s

[tex]\underset{v_{f}}{\rightarrow}[/tex] = final velocity of the baseball after collision = 56 [tex]\hat{j}[/tex] m/s

m = mass of the ball = 145 g = 0.145 kg

t = time of contact of the ball with the bat = 1.90 ms = 0.0019 s

[tex]\underset{F_{avg}}{\rightarrow}[/tex] = Average force vector

Using Impulse-change in momentum equation

[tex]\underset{F_{avg}}{\rightarrow}[/tex] t = m ([tex]\underset{v_{f}}{\rightarrow}[/tex] - [tex]\underset{v_{i}}{\rightarrow}[/tex] )

[tex]\underset{F_{avg}}{\rightarrow}[/tex] (0.0019) = (0.145) (56 [tex]\hat{j}[/tex] - 43 [tex]\hat{i}[/tex])

[tex]\underset{F_{avg}}{\rightarrow}[/tex] = (4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])

Answer:

a) 40.2 liters = 40200 cubic centimeters

b) 40.2 liters = 0.0402 cubic meters

Explanation:

Volume of bottle = 40.2 liters.

a)

      1 liter = 1000 cubic centimeter

      40.2 liters = 40.2 x 1000 = 40200 cubic centimeters

b)

      1 liter = 0.001 cubic meter

      40.2 liters = 40.2 x 0.001 = 0.0402 cubic meters

Answer:

[tex]\beta = 41.68°[/tex]

Explanation:

according to snell's law

[tex]\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }[/tex]

refractive index of water n_w is 1.33

refractive index of glass  n_g  is 1.5

[tex]sin\alpha = \frac{n_w}{n_g}* sin30[/tex]

[tex]sin\alpha = 0.443[/tex]

now applying snell's law between air and glass, so we have

[tex]\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}[/tex]

[tex]sin\beta = \frac{n_g}{n_a} sin\alpha[/tex]

[tex]\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha][/tex]

we know that [tex]sin\alpha = 0.443[/tex]

[tex]\beta = 41.68°[/tex]

Answer:

The pendulum clock does it loses 83.0304 seconds a day at the new location.

Explanation:

T1= 1 s

T2= ?

g1= 9.8135 m/s

g2= 9.7943 m/s

L=?

L= (T1/2π)²  * g1=

L= 0.24857m

T2= 2π * √(L/g2)

T2= 1.000961 s

ΔT= T2 - T1

ΔT = 0.000961 s

seconds lost a day = 24 * 3600 * ΔT

seconds lost a day= 24 * 3600 * 0.000961 s

seconds lost a day= 83.0304 s

A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity. The period of a pendulum is dependent on its length and the acceleration due to gravity. To calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds.

A pendulum clock loses accuracy when it is moved from one location to another due to changes in the acceleration due to gravity, not changes in the length of the pendulum. In this case, the clock was moved from a location where g = 9.8135 m/s2 to a new location where g = 9.7943 m/s2. The clock loses accuracy because the period of a pendulum is dependent on its length and the acceleration due to gravity.

The period of a pendulum can be calculated using the equation: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

So, to calculate how many seconds a day the clock loses at the new location, we can subtract the new period from the original period and convert it to seconds. Since the original period is 2.00000 s, we have:

ΔT = Toriginal - Tnew

ΔT = 2.00000 s - Tnew

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Answer:

0.4 times the radius of moon

Explanation:

gravity on moon is equal to the one sixth of gravity on earth.

g' = g / 6

where, g' is the gravity on moon and g be the gravity on earth.

As the earth shrinks, the mass of earth remains same.

The acceleration due to gravity is inversely proportional to the square of radius of planet.

g' ∝ 1/R'²   .....(1)

Where, R' is the radius of moon.

g ∝ 1/R²      ..... (2)

Where, R be the radius of earth.

Divide equation (1) by (2)

g / g' = R'² / R²

Put g' = g / 6

6 = R'² / R²

2.5 = R' / R

R = R' / 2.5 = 0.4 R'

Thus, the radius of earth should be 0.4 times the radius of moon.

Answer:

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

[tex]F = \frac{\mu_0 i_1 i_2}{2 \pi d}[/tex]

here we know that

[tex]F = 7.83 \times 10 N/m[/tex]

[tex]d = 7.0 cm = 0.07 m[/tex]

[tex]i_1 = 28 A[/tex]

now we will have

[tex]F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}[/tex]

[tex]7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}[/tex]

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction

Answer:

Time taken, t = 3 s

Explanation:

It is given that,

Initial velocity of the particle, u = 50 m/s

Final velocity, v = 20 m/s

Distance covered, s = 105 m

Firstly we will find the acceleration of the particle. It can be calculated using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]a=\dfrac{v^2-u^2}{2s}[/tex]

[tex]a=\dfrac{(20\ m/s)^2-(50\ m/s)^2}{2\times 105\ m}[/tex]

[tex]a=-10\ m/s^2[/tex]

So, the particle is decelerating at the rate of 10 m/s². Let t is the time taken for the particle to slow down. Using first equation of motion as :

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{20\ m/s-50\ m/s}{-10\ m/s^2}[/tex]

t = 3 s

So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.

Answer:

a)The parachutist in the air for 12.63 seconds.

b)The parachutist falls from a height of 293 meter.

Explanation:

Vertical motion of  parachutist:

 Initial speed, u = 0m/s

 Acceleration, a = 9.81 m/s²

 Displacement, s = 50 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 0² + 2 x 9.81 x 50

    v = 31.32 m/s

 Time taken for this

          31.32 = 0 + 9.81 x t

            t = 3.19 s

 After 50m we have

 Initial speed, u = 31.32m/s

 Acceleration, a = -2 m/s²

 Final speed , v = 3 m/s

 We have equation of motion, v² = u² + 2as

 Substituting

   3² = 31.32² - 2 x 2 x s

    s = 243 m

 Time taken for this

          3 = 31.32 - 2 x t

            t = 9.44 s

a) Total time = 3.19 +  9.44 = 12.63 s

    The parachutist in the air for 12.63 seconds.

b) Total height = 50 + 243 = 293 m

    The parachutist falls from a height of 293 meter.

Answer: (a) 1.065 N  (b) 2.13 N

Explanation:

According Newton's 2nd Law of Motion the force [tex]F[/tex] is defined as the variation of linear momentum [tex]p[/tex] in time:

[tex]F=\frac{dp}{dt}[/tex]  (1)

Where the linear momentum is:

[tex]p=mV[/tex]  (2) Being [tex]m[/tex] the mass and [tex]V[/tex] the velocity.

In the case of the rain drops, which initial velocity is [tex]V_{i}=15m/s[/tex] and final velocity is  [tex]V_{f}=0[/tex] (we are told the drops come to rest after striking the roof). The momentum of the drops [tex]p_{drops}[/tex] is:

[tex]p_{drops}=mV_{i}+mV_{f}[/tex]  (3)

If [tex]V_{f}=0[/tex], then:

[tex]p_{drops}=mV_{i}[/tex]  (4)

Now the force [tex]F_{drops}[/tex] exerted by the drops is:

[tex]F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}[/tex]  (5)

[tex]F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}[/tex]  (6)

At this point we know the mass of rain per second (mass rate) [tex]\frac{dm}{dt}=0.071 kg/s[/tex] and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

[tex]F_{drops}=\frac{dm}{dt}V_{i}[/tex]  (7)

[tex]F_{drops}=(0.071 kg/s)(15m/s)[/tex]  (8)

[tex]F_{drops}=1.065kg.m/s^{2}=1.065N[/tex]  (9) This is the force exerted by the rain drops on the roof of the car.

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum [tex]p_{hailstones}[/tex] is:

[tex]p_{hailstones}=mV_{i}+mV_{f}[/tex]   (9)  Being [tex]V_{i}=V_{f}[/tex]

[tex]p_{hailstones}=mV+mV=2mV[/tex]   (10)  

Deriving with respect to time to find the force [tex]F_{hailstones}[/tex] exerted by the hailstones:

[tex]F_{hailstones}=\frac{d}{dt}p_{hailstones}=\frac{d}{dt}(2mV)[/tex]   (10)  

[tex]F_{hailstones}=2\frac{d}{dt}(mV)=2(\frac{dm}{dt}V+m\frac{dV}{dt})[/tex]   (11)  

Assuming [tex]\frac{dV}{dt}=0[/tex]:

[tex]F_{hailstones}=2(\frac{dm}{dt}V)[/tex]   (12)  

[tex]F_{hailstones}=2(0.071 kg/s)(15m/s)[/tex]   (13)  

Finally:

[tex]F_{hailstones}=2.13kg.m/s^{2}=2.13N[/tex] (14)   This is the force exerted by the hailstones  

Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.

The average force exerted by the rain on the roof of a parked car can be calculated using the equation Force = mass × acceleration. If hailstones with the same mass and speed fall on the roof, the average force on the roof would be the same.

(a) To find the average force exerted by the rain on the roof, we can use the equation:

Force = mass × acceleration

The mass of rain per second striking the roof is given as 0.071 kg/s. Since the drops come to rest after striking the roof, the acceleration is equal to the initial velocity of the drops, which is 15 m/s. Therefore, the force is:

Force = 0.071 kg/s × 15 m/s = 1.065 N

(b) If hailstones with the same mass as the raindrops fall on the roof at the same rate and with the same speed, the average force on the roof would be the same as found in part (a). The mass and speed of the hailstones are the same as the raindrops, so the force exerted by the hailstones on the roof would also be 1.065 N.

Answer:

[tex]\lambda = 2.57 \times 10^{-11} m[/tex]

Explanation:

Average velocity of oxygen molecule at given temperature is

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

[tex]v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}[/tex]

[tex]v_{rms} = 483.4 m/s[/tex]

now for de Broglie wavelength we know that

[tex]\lambda = \frac{h}{mv}[/tex]

[tex]\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}[/tex]

[tex]\lambda = 2.57 \times 10^{-11} m[/tex]

Answer:

106.83

Explanation:

N = 332, l = 14 cm = 0.14 m, i = 0.88 A, B = 0.28 T

Let ur be the relative permeability

B = u0 x ur x n x i

0.28 = 4 x 3.14 x 10^-7 x ur x 332 x 0.88 / 0.14       ( n = N / l)

ur = 106.83

Answer:

0.012 m

Explanation:

m = mass of the marble = 0.0265 kg

M = mass of the pendulum = 0.250 kg

v = initial velocity of the marble before collision = 5.05 m/s

V = final velocity of marble-pendulum combination after the collision = ?

using conservation of momentum

m v = (m + M) V

(0.0265) (5.05) = (0.0265 + 0.250) V

V = 0.484 m/s

h = height gained by the marble-pendulum combination

Using conservation of energy

Potential energy gained by the combination = Kinetic energy of the combination just after collision

(m + M) gh = (0.5) (m + M) V²

gh = (0.5) V²

(9.8) h = (0.5) (0.484)²

h = 0.012 m

Using the concept of inelastic collison and principle of conservation of momentum, the height of swing made by the pendulum would be 0.012 meters

Collison is inelastic :

(0.0265 × 5.05) = (0.0265 + 0.250)v

0.133825 = 0.2765v

v = (0.133825 ÷ 0.2765)

v = 0.484 m/s

Final velocity after collision = 0.484 m/s

Assuming linear momentum is conserved :

0.5mv² = mgh

Mass cancels out

0.5v² = gh

0.5(0.484)² = 9.8h

9.8h = 0.1171

h = (0.1171) ÷ 9.8

h = 0.0119

Therefore, height of the swing made by the pendulum would be 0.012 meters.

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Answer:

0.25

Explanation:

A = area of each plate = 30 cm² = 30 x 10⁻⁴ m²

d = separation between the plates = 5 mm = 5 x 10⁻³ m

[tex]C_{air}[/tex] = Capacitance of capacitor when there is air between the plates

k = dielectric constant = 4

[tex]C_{dielectric}[/tex] = Capacitance of capacitor when there is dielectric between the plates

Capacitance of capacitor when there is air between the plates is given as

[tex]C_{air} = \frac{\epsilon _{o}A}{d}[/tex]                            eq-1

Capacitance of capacitor when there is dielectric between the plates is given as

[tex]C_{dielectric} = \frac{k \epsilon _{o}A}{d}[/tex]                            eq-2

Dividing eq-1 by eq-2

[tex]\frac{C_{air}}{C_{dielectric}}=\frac{\frac{\epsilon _{o}A}{d}}{\frac{k \epsilon _{o}A}{d}}[/tex]

[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{k}[/tex]

[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{4}[/tex]

[tex]\frac{C_{air}}{C_{dielectric}}=0.25[/tex]

Charge stored in the capacitor when there is air is given as

[tex]Q_{air}=C_{air}V[/tex]                             eq-3

Charge stored in the capacitor when there is dielectric is given as

[tex]Q_{dielectric}=C_{dielectric}V[/tex]                             eq-4

Dividing eq-3 by eq-4

[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}V}{C_{dielectric} V}[/tex]

[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}}{C_{dielectric}}[/tex]

[tex]\frac{Q_{air}}{Q_{dielectric}}=0.25[/tex]

Answer:

Option B is the correct answer.

Explanation:

Let us consider 40 meter above ground as origin.

Initial velocity = 17 m/s

Final velocity = 24 m/s

Acceleration = 9.81 m/s

We have equation of motion v² = u² + 2as

Substituting

         24² = 17² + 2 x 9.81 x s

           s = 14.63 m

Distance traveled by rock = 14.63 m down.

Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m

Option B is the correct answer.

Answer:

(a) No. of protons = atomic number = 7

No. of electrons = no. of protons = 7

No. of neutrons = mass no. - atomic no. = 6

(b) No. of protons = atomic number = 16

No. of electrons = no. of protons = 16

No. of neutrons = mass no. - atomic no. = 19

(c) No. of protons = atomic number = 29

No. of electrons = no. of protons = 29

No. of neutrons = mass no. - atomic no. = 34

(d) No. of protons = atomic number = 38

No. of electrons = no. of protons = 38

No. of neutrons = mass no. - atomic no. = 51  

Explanation:

No. of protons = No. of electrons = atomic number

No. of neutrons = Mass no. - atomic no.

(a) 13 N 7

Here atomic number is 7, mass number is 13

No. of protons = atomic number = 7

No. of electrons = no. of protons = 7

No. of neutrons = mass no. - atomic no. = 13 - 7 = 6

(b) 35 S 16

Here atomic number is 16, mass number is 35

No. of protons = atomic number = 16

No. of electrons = no. of protons = 16

No. of neutrons = mass no. - atomic no. = 35 - 16 = 19

(c) 63 Cu 29

Here atomic number is 29, mass number is 63

No. of protons = atomic number = 29

No. of electrons = no. of protons = 29

No. of neutrons = mass no. - atomic no. = 63 - 29 = 34

(d) 89 Sr 38

Here atomic number is 38, mass number is 89

No. of protons = atomic number = 38

No. of electrons = no. of protons = 38

No. of neutrons = mass no. - atomic no. = 89 - 38 = 51

Final answer:

The isotopes ¹N⁷, ³⁵S¹⁶, ⁶³Cu₂₉, and ₈₉Sr₃₈ have 7, 16, 29, and 38 protons, respectively, with neutrons calculated by subtracting the number of protons from the mass number. Lastly, the number of electrons is equal to the number of protons in a neutral atom.

Explanation:

The number of protons, neutrons, and electrons for each isotope can be determined using the atomic number, mass number, and charge. The atomic number represents the number of protons (and also the number of electrons in a neutral atom), the mass number represents the sum of protons and neutrons, and the charge indicates how many electrons have been lost or gained.

Each isotope comprises an equal number of protons and electrons in their neutral states. Neutrons are the difference between the mass number and the number of protons.

Explanation:

The motion of a particle is defined by the relation as:

[tex]x=2t^3-9t^2+12t+10[/tex]........(1)

Differentiating equation (1) wrt t we get:

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=\dfrac{d(2t^3-9t^2+12t+10)}{dt}[/tex]

[tex]v=6t^2-18t+12[/tex]............(2)

When v = 2 ft/s

[tex]2=6t^2-18t+12[/tex]

[tex]6t^2-18t+10=0[/tex]

t₁ = 2.26 s

and t₂ = 0.73 s

Put the value of t₁ in equation (1) as :

[tex]x=2(2.26)^3-9(2.26)^2+12(2.26)+10[/tex]

x₁ = 14.23 ft

Put the value of t₂ in equation (1) as :

[tex]x=2(0.73)^3-9(0.73)^2+12(0.73)+10[/tex]

x₁ = 14.74 ft

For acceleration differentiate equation (2) wrt t as :

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]a=\dfrac{d(6t^2-18t+12)}{dt}[/tex]

a = 12 t - 18.........(3)

Put t₁ and t₂ in equation 3 one by one as :

[tex]a_1=12(2.26)-18=9.12\ ft/s^2[/tex]

[tex]a_2=12(0.73)-18=-9.24\ ft/s^2[/tex]

Hence, this is the required solution.

The time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.

When an object is observed to move in a straight line, then the motion of the object is known as linear motion. It can be referenced as motion in one dimension.

Given data:

The position of the particle is, [tex]x = 2t^{3}-9t^{2}+12t+10[/tex].

The velocity of the particle is, [tex]v = 2.00 \;\rm ft/s[/tex].

We know that velocity can be obtained by differentiating the position. Then,

[tex]v =\dfrac{dx}{dt}\\\\v = \dfrac{d(2t^{3}-9t^{2}+12t+10)}{dt}\\\\v=6t^{2}-18t+12[/tex]

Now when v = 2.00 ft/s. The time is,

[tex]2 = 6t^{2}-18t+12\\\\6t^{2}-18t+10=0\\\\t = 2.26 \;\rm s[/tex]

Now, the position of the particle is,

[tex]x = 2t^{3}-9t^{2}+12t+10\\\\x = (2 \times 2.26^{3})-(9 \times 2.26^{2})+(12 \times 2.26)+10\\\\x =14.23\;\rm ft[/tex]

And the magnitude of the acceleration is calculated by differentiating the velocity as,

[tex]a = \dfrac{dv}{dt}\\\\a = \dfrac{d(6t^{2}-18t+12)}{dt}\\\\a=12t-18[/tex]

Substituting t = 2.26 s as,

[tex]a = 12(2.26)-18\\\\a=9.12 \;\rm ft/s^{2}[/tex]

Thus, we can conclude that the time taken by the particle is 2.26 s and the magnitude of acceleration and position of the particle at the obtained instant are [tex]9.12 \;\rm ft/s^{2}[/tex] and 14.23 ft respectively.

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Answer:

He made it, he across safely the river without falling ing.

Explanation:

Vt= 7.8 m/s

r= 11m

m= 80.5 kg

Vt= ω * r

ω= 0.71 rad/s

ac= ω² * r

ac= 5.54 m/s²

F= m * ac

F= 445.97 N  < 1,000 N

Answer:

The electric potential difference is 32500 volt.

Explanation:

Given that,

Charge[tex]q=2\times10^{-4}C[/tex]

Energy = 6.5 J

We need to calculate the electric potential difference

Potential difference :

Potential difference is equal to the energy divide by charge.

Using formula of potential difference

[tex]V=\dfrac{E}{Q}[/tex]

[tex]V=\dfrac{6.5}{2\times10^{-4}}[/tex]

[tex]V=32500\ volt[/tex]

Hence, The electric potential difference is 32500 volt.

The electric potential difference is 32,500 volts.

To determine the electric potential difference, we can use the formula:

V = ΔE / q

where V is the electric potential difference, ΔE is the energy and q is the test charge. In this case, the energy given is 6.5J and the test charge is 2 x 10-4C. Plugging these values into the formula, we get:

V = 6.5J / (2 x 10-4C)

V = 32,500V

Therefore, the electric potential difference is 32,500 volts.

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Answer:

P₁ = 192 W and P₂ = 384 W

Explanation:

It is given that,

Resistor 1, [tex]R_1=12\ \Omega[/tex]

Resistor 2, [tex]R_1=6\ \Omega[/tex]

Voltage, V = 48 V

Power dissipated by resistor 1 is given by :

[tex]P_1=\dfrac{V^2}{R_1}[/tex]

[tex]P_1=\dfrac{(48)^2}{12}[/tex]

P₁ = 192 watts

Power dissipated by resistor 2 is given by :

[tex]P_2=\dfrac{V^2}{R_1}[/tex]

[tex]P_2=\dfrac{(48)^2}{6}[/tex]

P₂ = 384 watts

So, the power dissipated by both the resistors is 192 watts and 384 watts respectively. Hence, this is the required solution.

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time [tex]t_1[/tex], the acceleration vector has direction [tex]\theta_1[/tex] such that

[tex]\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ[/tex]

which indicates the particle is situated at a point on the lower left half of the circle, while at time [tex]t_2[/tex] the acceleration has direction [tex]\theta_2[/tex] such that

[tex]\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ[/tex]

which indicates the particle lies on the upper left half of the circle.

Notice that [tex]\theta_1-\theta_2=90^\circ[/tex]. That is, the measure of the major arc between the particle's positions at [tex]t_1[/tex] and [tex]t_2[/tex] is 270 degrees, which means that [tex]t_2-t_1[/tex] is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

[tex]\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}[/tex]

where [tex]R[/tex] is the radius of the circle and [tex]T[/tex] is the period. We have

[tex]t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s[/tex]

and the magnitude of the particle's acceleration toward the center of the circle is

[tex]\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}[/tex]

So we find that the path has a radius [tex]R[/tex] of

[tex]7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}[/tex]

The acceleration of a particle moving in a circular path with constant speed is given by the centripetal acceleration. Since the velocity is not given, the radius cannot be determined with this information alone.

In circular motion, the acceleration of a particle is given by the centripetal acceleration, which is equal to the square of the speed divided by the radius of the circular path (v^2/r). As the particle moves at a constant speed, we know that the magnitude of acceleration (which gives us the centripetal acceleration) remains the same at both times t1 and t2. Therefore, we can calculate the magnitude of acceleration at both points and set them equal to each other. This will allow us to solve for the radius of the path.

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Explanation:

It is given that,

Initial speed, v₁ = 14 m/s

Initial force, F₁ = 130 N

We need to find the total horizontal force (F₂) on the driver if the speed on the same curve is 18.0 m/s instead, v₂ = 18 m/s

The centripetal force is given by :

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

[tex]\dfrac{F_1}{F_2}=\dfrac{mv_1^2/r}{mv_2^2/r}[/tex]

[tex]\dfrac{F_1}{F_2}=\dfrac{v_1^2}{v_2^2}[/tex]

[tex]F_2=\dfrac{v_2^2\times F_1}{v_1^2}[/tex]

[tex]F_2=\dfrac{(18\ m/s)^2\times 130\ N}{(14\ m/s)^2}[/tex]

[tex]F_2=214.8\ N[/tex]

So, if the speed is 18 m/s, then the horizontal force acting on the car is 214.8 N. Hence, this is the required solution.

Final answer:

The total horizontal force on the driver when the car's speed increases to 18.0 m/s can be found using the proportionality between the centripetal force and the square of the velocity. By comparing the initial and final conditions, we can calculate the new total horizontal force required for the increased speed.

Explanation:

The question involves finding the total horizontal force on a driver when the speed of the car increases on the same curve. The horizontal force experiences by the driver in a curved path, which is part of a horizontal circle, can be determined by using the formula for centripetal force (Fc = mv²/r). We know from the initial condition that the horizontal force is 130 N when the car is traveling at 14.0 m/s. Since the force depends on the square of the velocity, we can use proportions to find the new force when the speed increases to 18.0 m/s.

Thus, (F2/F1) = (v₂²/v₁²). Substituting the given values, we get (F2/130 N) = (18.0 m/s)² / (14.0 m/s)². Calculating this provides us with the new force F2 which represents the total horizontal force on the driver at the higher speed of 18.0 m/s.

Answer:

Volumetric flow rate = 0.024 m³/s  

Explanation:

Volumetric flow rate = Discharge = Q

We have expression for discharge, Q = Area ( A) x Velocity (v)

Velocity , v = 3 m/s

[tex]\texttt{Area, A}=\frac{\pi d^2}{4}=\frac{\pi \times (10\times 10^{-2})^2}{4}=7.85\times 10^{-3}m^2[/tex]

Substituting

  Discharge, Q =Av = 7.85 x 10⁻³ x 3 = 0.024 m³/s  

Volumetric flow rate = 0.024 m³/s  

To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), multiply by 694.7. To convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), multiply by 11,545.

To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), we can use the following conversion factors:

Using these conversion factors, we can convert the thermal conductivity value as follows:

0.3 Btu/(h ft°F) = (0.3 * 1055.06 J)/(h * 0.3048 m * 0.5556 °C) = 694.7 W/(m °C)

Similarly, to convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), we can use the following conversion factors:

Using these conversion factors, we can convert the surface heat transfer coefficient value as follows:

105 Btu/(h ft2 oF) = (105 * 1055.06 J)/(h * 0.0929 m2 * 0.5556 °C) = 11,545 W/(m2 °C)

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The conversion of the thermal conductivity value yields approximately 0.52 W/(m°C), and the surface heat transfer coefficient conversion yields approximately 596.15 W/(m²°C). Conversion factors for BTU to J, ft to m, and °F to °C are critical in these calculations.

To convert thermal conductivity from BTU/(h ft°F) to W/(m°C):

To convert surface heat transfer coefficient from BTU/(h ft²°F) to W/(m²°C):

Final answer:

The weight of the piece when made of lead is 992.6 lb. The weight of the piece when made of aluminum is 236.6 lb. By using aluminum instead of lead, a weight of 756 lb is saved.

Explanation:

(a) Weight of the piece when it is made of lead:

To find the weight of the piece when it is made of lead, we need to know the density of lead. The density of lead is approximately 709 lb/ft³. We can use the formula:

Weight = Density x Volume

Given that the volume of the piece is 1.4 ft³, the weight of the piece when it is made of lead can be calculated as:

Weight = 709 lb/ft³ x 1.4 ft³ = 992.6 lb

(b) Weight of the piece when made of aluminum, and weight saved:

To find the weight of the piece when it is made of aluminum, we need to know the density of aluminum. The density of aluminum is approximately 169 lb/ft³. Using the same formula as before:

Weight = Density x Volume

Given that the volume of the piece is still 1.4 ft³, the weight of the piece when it is made of aluminum can be calculated as:

Weight = 169 lb/ft³ x 1.4 ft³ = 236.6 lb

The weight saved by using aluminum instead of lead can be determined by subtracting the weight of the aluminum piece from the weight of the lead piece:

Weight saved = Weight of lead piece - Weight of aluminum piece

Weight saved = 992.6 lb - 236.6 lb = 756 lb

Answer:

[tex]R_{79} = 28.91 OHM[/tex]

Explanation:

Resistance ca be determine by using following formula

[tex]R = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]

where

R = Conductor resistance = 23.7Ω

Rref = conductor reistance at reference temperature,

α = temperature coefficient of resistance for material, for copper 40.41*10^{-4}

T = Conductor temperature in Celcius =  20.3°C

Tref = reference temperature at which α  is specified.

[tex]23.7 = R_{ref}\left [ 1+ 40.41*10^{-4} (20.3) \right ][/tex][tex]R_{ref} =21.92 OHM[/tex]

now for 79 degree celcius

[tex]R_{79} = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]

[tex]R_{79} =21.92\left [ 1+ 40.41*10^{-4} (79) \right ][/tex]

[tex]R_{79} = 28.91 OHM[/tex]