Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 18.5 kg and an initial velocity of v0A = 8.15 m/s, due east. Object B, however, has a mass of mB = 30.5 kg and an initial velocity of v0B = 5.00 m/s, due north. Find the magnitude of the final velocity of the two-object system after the collision.

Answer:

[tex]\Delta p=(0,3.10)kg*m/s\\[/tex]  

Explanation:

Momentum change:

[tex]\Delta p=p_{f}-p_{o}\\[/tex]   :  vector

p=mv

[tex]p_{o}=(p_{ox, p_{oy}}}=(m*v*sin(\theta),-m*v*cos(\theta) )\\[/tex]   : the ball move downward with an angle theta to the vertical

[tex]p_{f}=(p_{fx, p_{fy}}}=(m*v*sin(\theta),+m*v*cos(\theta) )\\[/tex]     :the ball move upward with the same angle theta to the vertical, with same speed

So:

[tex]\Delta p=p_{f}-p_{o}=(0,2m*v*cos(\theta))=(0,2*0.609*3*cos(32))=(0,3.10)kg*m/s\\[/tex]  

Answer:

The distance is [tex]9.57\times10^{13}\ m[/tex]

Explanation:

Given that,

Diameter of telescope d= 5.08 m

Diameter of sun spot y= 10000 mi

[tex]y =1609.3\times10^{4}\ m[/tex]

We need to calculate the distance

Using formula of distance

[tex]y =\dfrac{1.22\lambda D}{d}[/tex]

[tex]D=\dfrac{d y}{1.22\times\lambda}[/tex]

Put the value into the formula

[tex]D=\dfrac{5.08\times1609.3\times10^{4}}{1.22\times700\times10^{-9}}[/tex]

[tex]D=9.57\times10^{13}\ m[/tex]

Hence, The distance is [tex]9.57\times10^{13}\ m[/tex]

Final answer:

The most distant star on which the Hale Telescope could resolve a sunspot of this size is approximately 73 trillion km away.

Explanation:

To determine the most distant star that this telescope could resolve a sunspot on, we need to calculate the angular resolution of the telescope. The angular resolution of a telescope is given by the formula θ = 1.22 * (λ / D), where θ is the angular resolution, λ is the wavelength of light, and D is the diameter of the telescope's mirror.

To resolve the diameter of a sunspot, we can use the approximate diameter of a large sunspot of 10,000 miles. Converting this to kilometers, we get 16,093.44 km. Assuming an average wavelength of visible light of 550 nm, we can calculate the angular resolution as follows:

θ = 1.22 × (550 nm / 5.08 meters) = 0.013 arc seconds

Now, we need to determine the distance at which this angular resolution corresponds to a sunspot diameter of 16,093.44 km. We can use the small angle formula to calculate the distance:

D = diameter / tan(θ)

D = 16,093.44 km / tan(0.013 arc seconds) = 73,259,925,487,382 km

Therefore, the most distant star on which the Hale Telescope could resolve a sunspot of this size is approximately 73 trillion km away.

Answer:

(a) Projectile B will travel 4 times as far as projectile A prior to landing

Explanation:

Initial velocity = v

Angle at which the projectile is shot at = θ

g = Acceleration due to gravity

Range of a projectile is given by

[tex]R=\frac {v^{2}\sin 2\theta}{g}[/tex]

When Initial velocity = v

[tex]R_A=\frac{v^{2}\sin 2\theta}{g}[/tex]

When Initial velocity = 2v

[tex]R_B=\frac{(2v)^{2}\sin 2\theta}{g}\\\Rightarrow R_B=\frac{4v^2\sin 2\theta}{g}[/tex]

Dividing the equtions, we get

[tex]\frac{R_A}{R_B}=\frac{\frac{v^{2}\sin 2\theta}{g}}{\frac{4v^2\sin 2\theta}{g}}[/tex]

Here, the angle at which the projectiles are fired at are equal.

[tex]\frac{R_A}{R_B}=\frac{1}{4}\\\Rightarrow R_B=4R_A[/tex]

Hence, projectile B will travel 4 times as far as projectile A prior to landing

To compare the ranges of the two projectiles, we can use the fact that the horizontal range of a projectile launched with initial speed [tex]\( v_0 \)[/tex]at an angle [tex]\( \theta \)[/tex] with respect to the horizontal is given by:

[tex]\[ R = \frac{v_0^2 \sin(2\theta)}{g} \][/tex]

Where:

-  R  is the range,

- [tex]\( v_0 \)[/tex] is the initial speed of the projectile,

- [tex]\( \theta \)[/tex] is the launch angle, and

- g is the acceleration due to gravity.

Both projectiles are launched at the same angle with respect to the ground. Since the launch angle is the same for both projectiles, we can compare their ranges by comparing their initial speeds.

Let's denote the range of projectile A as [tex]\( R_A \)[/tex] and the range of projectile B as [tex]\( R_B \).[/tex]

For projectile A:

[tex]\[ R_A = \frac{v^2 \sin(2\theta)}{g} \][/tex]

For projectile B:

[tex]\[ R_B = \frac{(2v)^2 \sin(2\theta)}{g} = 4 \times \frac{v^2 \sin(2\theta)}{g} = 4R_A \][/tex]

So, the range of projectile B is four times the range of projectile A.

Therefore, the correct answer is:

(a) Projectile B will travel 4 times as far as projectile A prior to landing.

Explanation:

The nucleus of an atom can be modeled as several protons and neutrons closely packed together.

Mass of the particle, [tex]m=1.67\times 10^{-27}\ kg[/tex]

Radius of the particle, [tex]R=10^{-15}\ m[/tex]

(a) The density of the nucleus of an atom is given by mass per unit area of the particle. Mathematically, it is given by :

[tex]d=\dfrac{m}{V}[/tex], V is the volume of the particle

[tex]d=\dfrac{m}{(4/3)\pi r^3}[/tex]

[tex]d=\dfrac{1.67\times 10^{-27}}{(4/3)\pi (10^{-15})^3}[/tex]

[tex]d=3.98\times 10^{17}\ kg/m^3[/tex]

So, the density of the nucleus of an atom is [tex]3.98\times 10^{17}\ kg/m^3[/tex].

(b) Density of iron, [tex]d'=7874\ kg/m^3[/tex]

Taking ratio of the density of nucleus of an atom and the density of iron as :

[tex]\dfrac{d}{d'}=\dfrac{3.98\times 10^{17}}{7874}[/tex]

[tex]\dfrac{d}{d'}=5.05\times 10^{13}[/tex]

[tex]d=5.05\times 10^{13}\ d'[/tex]

So, the density of the nucleus of an atom is [tex]5.05\times 10^{13}[/tex] times greater than the density of iron. Hence, this is the required solution.

Answer:

zero, acceleration due to gravity = 9.8 m/s^2

Explanation:

When an object throws vertically upwards, the acceleration acting on the object is acceleration due to gravity which is acting in vertically downwards direction.

As the object moves upwards, its velocity goes on decreasing because the direction of velocity and the acceleration both are opposite to each other. At maximum height, the velocity of object becomes zero and then object starts moving in downwards direction.

Thus, the value of instantaneous velocity at maximum height is zero but the vale of acceleration is acceleration due to gravity which is acting vertically downward direction.

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

[tex]p_i=mv+0=mv[/tex]

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

[tex]p_f=mv'+MV'[/tex]

Equating initial and the final momenta we get

[tex]mv=mv'+MV'\\\\m(v-v')=MV'.....i[/tex]

Now since the surface is frictionless thus the energy is also conserved thus

[tex]E_i=\frac{1}{2}mv^2[/tex]

Similarly the final energy becomes

[tex]E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2[/tex]\

Equating initial and final energies we get

[tex]\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)[/tex]

Solving i and ii we get

[tex]v+v'=V'[/tex]

Using this in equation i we get

[tex]v'=\frac{v(m-M)}{(M-m)}=-v[/tex]

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

Answer:

216480000 fathoms

Explanation:

1 fathom = 6 feet

[tex]1\ feet=\frac{1}{6}\ fathom[/tex]

Distance from Earth to the Moon = 246000 miles

Converting to feet

1 mile = 5280 feet

246000 miles = 1298880000 feet

Convert to fathom

[tex]1\ feet=\frac{1}{6}\ fathom\\\Rightarrow 1298880000\ feet=\frac{1298880000}{6}=216480000\ fathom[/tex]

So, the distance between Earth and Moon is 216480000 fathoms

Answer:

Vmax = 12.21m/s

[tex]a = 4.07m/s^{2}[/tex]

Explanation:

For the first 3 seconds:

[tex]V_{3}=V_{o}+a*t=0+a*3=3a[/tex]   This is the maximum speed. But we need acceleration.

[tex]X_{3}=V_{o}*t+\frac{a*t^{2}}{2} =\frac{9*a}{2}[/tex]

For the other 6.69s with constant speed:

[tex]X_{t}=100=X_{3}+V_{3}*t[/tex]

[tex]100=\frac{9*a}{2} +3*a*6.69[/tex]   Solving for a:

[tex]a = 4.07m/s^{2}[/tex]  Now we replace this value on [tex]V_{3}=3a[/tex]:

[tex]V_{3}=V_{max}=12.21m/s[/tex]

Answer:

Explanation:

Spring constant k = 5N/m

I ) x = 3 m means , spring is stretched by 3 m  

Restoring force by spring = kx = 5 x 3 = 15 N

II )  When mass is touching the wall, extension in spring = 1 m

Force by spring on the body

= 1 x 5 = 5 N .

iii ) . It is touching the wall , x =  1 m

Stored energy in the spring = 1/2 k x² = .5 x 5 x 1 x 1

= 2.5 J

When x = 3 , energy stored in it

Potential energy stored in it = 1/2 k x²

= .5 x 5 x 3 x 3

= 22.5 J

Increase in stored energy = 22.5 - 2.5

20 J

This must be the work done to stretch it from 1 m to 3 m .

Answer:2.83 m/s

Explanation:

Given

Object starts at x=-13 m at t=0 s

object takes 18 s to travel 51 m with constant velocity

i.e. there is no acceleration

and [tex]distance =speed\times times[/tex]

[tex]51=v\times 18[/tex]

v=2.83 m/s

A molecule of hydrogen moving at a speed of 115 cm/s would take approximately 79.5 seconds to travel the length of a football field (100 yd).

To compute the time it would take a molecule of hydrogen to move the length of a football field, you would use the formula time = distance/speed. However, we must first convert the length of the football field from yards to centimeters for consistency.

One yard is approximately 91.44 cm, so a football field which is 100 yards is about 9144 cm. Using the given speed of a hydrogen molecule which is 115 cm/s, the time it would take can be computed as follows: time = 9144 cm / 115 cm/s = approx. 79.5 seconds.

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Answer:

The magnitude of force is [tex]4.26\times 10^{- 6} N[/tex]

Solution:

As per the question:

The strength of Electric field due west at a certain point, [tex]\vec{E_{w}} = 710,000 N/C[/tex]

Charge, Q = - 6 C

Now, the force acting on the charge Q in the electric field is given by:

[tex]\vec{F} = Q\vec{E_{w}}[/tex]

[tex]\vec{F} = -6\times 710,000 = - 4.26\times 10^{- 6} N[/tex]

Here, the negative sign indicates that the force acting is opposite in direction.

Answer:

93.29 gallons

Explanation:

Given:

Number of solar panels = 3

Area of each solar panel = 1 m × 2 m = 2 m²

Total area of solar panels = 3 × 2 = 6 m²

Time = 4 hrs = 4 × 60 × 60 = 14400 seconds

Change in temperature, ΔT = 120° C - 40° C = 80° C

Now,

the solar power received on the Earth = 1368 W/m²

Thus,

The Heat energy received = Power × Area × Time

or

The Heat energy received = 1368 × 6 × 14400 =  118195200 J

Also,

Heat = mCΔT

where, C is the specific heat of the water

m is the mass of the water = 4.184 J/g.C

thus,

118195200 J = m × 4.184 × 80

or

mass of water that can be heated, m = 353116.63 grams = 353.116 kg

Also,

1 gallon of water = 3.785 kg

thus,

1 kg of water = 0.2642 gallons

Hence,

353.116 kg of water = 93.29 gallons

i.e 93.29 gallons of water can be heated

To determine how many gallons of water can be heated from 40°C to 120°C by three solar panels, we must calculate the energy absorbed by the panels using insolation data and then apply the specific heat capacity of water. The actual insolation value is necessary for the precise calculation, which was not provided in the question.

The amount of water that can be heated from 40°C to 120°C each day by solar panels can be calculated using principles from physics, specifically thermodynamics and energy transfer. First, we need to determine the energy incident on the solar panels. Assuming normally incident sunlight for 4 hours and a total solar panel area of 6 m² (as there are three 1 m x 2 m panels), we can calculate the energy absorbed.

Next, we use the specific heat capacity equation to find how much water this energy can heat from 40°C to 120°C. The specific heat capacity of water is approximately 4.18 J/g°C. Afterwards, we'll convert this amount of water from liters (or kilograms, as 1 L of water is approximately 1 kg) into gallons for the answer.

To complete this calculation, we would need to know the actual insolation in the specific location in terms of energy per unit area per unit time (e.g., kW/m²), which is not provided in the question or the reference material. But, if we assume insolation similar to the examples given in the reference material, we could use that to approximate the answer.

Answer:29.627 m

Explanation:

Given

Initial velocity of life preserver(u) is 1.6 m/s

it takes 2.3 s to reach the water

using equation of motion

v=u+at

[tex]v=1.6+9.81\times 2.3[/tex]

v=24.163 m/s

Let s be the height of life preserver

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

[tex]24.163^2-1.6^2=2\times 9.81\times s[/tex]

[tex]s=\frac{581.29}{2\times 9.81}[/tex]

s=29.627 m

Answer:

Case I: 12.617 L/s

Case II: 161.406 cubic meters per hour

Case III: 1.062 Pound inches

Explanation:

Given:

Assumptions:

Case I:

[tex]Speed\ of\ water\ flow = 200 \dfrac{gallon}{min}\\\Rightarrow V_{water} = 200\times \dfrac{3.785\ L}{60\ s}\\\Rightarrow V_{water} = 12.617\ L/s[/tex]

Case II:

[tex]Speed\ of\ air\ blow = 95 \dfrac{ft^3}{min}\\\Rightarrow V_{air} = 95\times \dfrac{(0.3048\ m)^3}{\dfrac{1}{60}\ h}\\\Rightarrow V_{air} = 95\times (0.3048)^3\times 60\ m^3/h\\\Rightarrow V_{air} = 161.406\ m^3/h[/tex]

Case III:

[tex]Measure\ of\ torque = 12\ N cm\\\Rightarrow \tau = 12\times (0.2248\ lb)\times \dfrac{1}{2.54}\ in\\ \Rightarrow \tau = 1.062\ lb in[/tex]

Answer:

The total electric potential at mid way due to 'q' is [tex]\frac{q}{4\pi\epsilon_{o}d}[/tex]

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say)  = d

The electric potential at a distance d due to 'Q' is:

[tex]V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}[/tex]

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

[tex]V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}[/tex]

Similar is the case with plate B:

[tex]V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}[/tex]

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

[tex]V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}[/tex]

[tex]V_{total} = \frac{q}{4\pi\epsilon_{o}d}[/tex]

Now,

The Electric field due to charge Q at a distance is given by:

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

Now, if the charge q is mid way between the field, then distance is [tex]\frac{d}{2}[/tex].

Electric Field at plate A, [tex]\vec{E_{A}}[/tex] at midway due to charge q:

[tex]\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}[/tex]

Similarly, for plate B:

[tex]\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}[/tex]

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

Answer:

a) 27.2 rad/min

b) 260 rev/h

Explanation:

The passenger is traveling at 9 mph, this is the tangential speed.

The relation between tangential speed and angular speed is:

v = r * w

Where

v: tangential speed

r: radius

w: angular speed

Also, the radius is

r = d/2

d is the diameter

Therefore:

v = (d * w)/2

Rearranging:

w = 2*v/d

w = (2*9 mile/h)/(58 feet)

We need to convert the feet to miles

w = (2*9 mile/h)/(0.011 miles) = 1636 rad/h

We divide this by 60 to get it in radians per minute

w = 1636/60 = 27.2 rad/min

Now the angular speed is in radians, to get revolutions we have to divide by 2π

n = v/(π*d)

n = (9 mile/h)/(π*0.011 mile) = 260 rev/h

The angular speed of a Ferris wheel with a 58-foot diameter, while carrying a passenger traveling at a speed of 9 miles per hour, is approximately 27.31 radians per minute. This Ferris wheel makes approximately 259 revolutions per hour.

To solve this problem, we first need to convert the linear speed from miles per hour to feet per minute, as the unit of the Ferris wheel’s diameter is in feet. One mile is equivalent to 5280 feet, and one hour is 60 minutes. Therefore, the passenger's speed in feet per minute (ft/min) is 9 miles/hour x 5280 feet/mile ÷ 60 minutes/hour = 792 ft/min.

(a) The angular speed in radians per minute can be found by dividing the linear speed by the radius of the wheel (which is half of the diameter). So, the wheel’s radius is 58 feet ÷ 2 = 29 feet, and thus, the angular speed is 792 ft/min ÷ 29 feet = 27.31 rad/min.

(b) The number of revolutions per hour is found by dividing the linear speed by the circumference of the wheel (which is the diameter × π). Therefore, the wheel's circumference is 58 feet x π. Consequently, the number of revolutions per hour is 792 ft/min x 60 min/hour ÷ (58 feet x π) ≈ 259 revolutions per hour, when rounded to the nearest whole number.

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Answer:

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance =  100 cm

v = Image distance

f = Focal length = -23.9 cm (concave lens)

[tex]h_u[/tex]= Object height = 2.1 cm

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-23.9}-\frac{1}{100}\\\Rightarrow \frac{1}{v}=\frac{-1239}{23900} \\\Rightarrow v=\frac{-23900}{1239}=-19.29\ cm[/tex]

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-19.29}{100}\\\Rightarrow m=0.1929[/tex]

[tex]m=\frac{h_v}{h_u}\\\Rightarrow 0.1929=\frac{h_v}{2.1}\\\Rightarrow h_v=0.1929\times 2.1=0.40509\ cm[/tex]

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

uranium atom = singly ionized

iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron  distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × [tex]10^{-19}[/tex] C

and charge on iron will be q3 = 2 × 1.602 × [tex]10^{-19}[/tex] C

so charge on electron is q1 =  - 1.602 × [tex]10^{-19}[/tex] C

and we know F = [tex]k\frac{q*q}{r^{2} }[/tex]  

so now by equilibrium

Fu = Fi

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

put here k = [tex]9*10^{9}[/tex] and find r

[tex]9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }[/tex]  =  [tex]9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }[/tex]

[tex]\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}[/tex]

r = 18.27 nm

distance r from the uranium atom is 18.27 nm

Answer:

a)Q= + 0.71 nC , For the resultant electric field at the origin to be 45.0 N/C in the +x direction

b)Q= -2.83nC ,for the resultant electric field at the origin to be 45.0 N/C in the −x direction

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻9 C

Data

k = 8.99*10⁹ N×m²/C²

q₁ =+5.45nC = 3*10⁻⁹C

d₁ =1.35 m

d₂ = 0.595m

a)Problem development : sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the +x direction

We make the algebraic sum of fields at at the origin :

[tex]E_{o} =E_{q} +E_{Q}[/tex] Equation (1)

[tex]E_{q} =\frac{k*q_{1} }{d_{1}{2}  }[/tex]

Calculation of E(q)

[tex]E_{q} =\frac{8.99*10^{9} *5.45*10^{-9} }{1,35^{2} }[/tex]

[tex]E_{q} =26.88\frac{N}{C}[/tex] : in the +x direction .As the charge is negative, the field enters the charge

We replace [tex]E_{o}[/tex] and [tex]E_{q}[/tex] in the equation (1)

[tex]45=26.88+E_{Q}[/tex]

[tex]E_{Q} =45-26.88[/tex]

[tex]E_{Q} = 18.12 N/C[/tex] : in the +x direction .

Sign and magnitude of Q

Q must be positive for the field to abandon the load in the +x

[tex]E_{Q} =\frac{k*Q}{d_{2}^{2} }[/tex]

[tex]18.12=\frac{8.99*10^{9}*Q }{0.595^{2} }[/tex]

[tex]Q=\frac{18.12*0.595^{2} }{8.99*10^{9} }[/tex]

Q=0.71*10⁻⁹ C =0.71 nC

b)Sign and magnitude of Q for the resultant electric field at the origin to be 45.0 N/C in the −x direction

We make the algebraic sum of fields at at the origin :

[tex]E_{o} =E_{q} +E_{Q}[/tex]

[tex]-45=26.88+E_{Q}[/tex]

[tex]-71.88=E_{Q}[/tex]

[tex]71.88=\frac{8.99*10^{9} *Q}{0.595^{2} }[/tex]

Q= 2.83*10⁻⁹ C

Q= -2.83nC

Q must be negative for the field to enters the charge in the −x direction

The magnitude and sign of Q is given by the required magnitude and

sign of the charge at the origin due to the sum of the charges.

Responses:

The charge at the origin is given as follows;

When the charge at the origin is 45.0 N/C, we have;

[tex]45 = \mathbf{\dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} + \dfrac{8.99 \times 10^{9} \times Q} {(-0.595)^2}}[/tex]

Which gives;

[tex]Q = \dfrac{\left(45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx \mathbf{2.83 \times 10^{-9}}[/tex]

When the charge at the origin is [tex]E_0[/tex] = 45 N/C, we have;

When the charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;

[tex]Q = \dfrac{\left(-45 - \dfrac{8.99 \times 10^{9} \times -5.45 \times ^{-9} }{1.35^2} \right) \times (-0.595)^2}{8.99 \times 10^{9} } \approx -7.134 \times 10^{-10}[/tex]

Therefore;

The charge at the origin is [tex]E_0[/tex] = 45 N/C in the -x direction, we have;

Learn more about electric field strength here:

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Explanation:

The maximum mass that can be hung from a string, m = 10 kg

Length of the string, l = 2 m

Mass of the object, m = 0.5 kg

Let v is the maximum speed that the mass can attain under these conditions without the string breaking. If T is the maximum tension in the string. So,

[tex]T_{max}=mg[/tex]

[tex]T_{max}=10\times 9.8=98\ N[/tex]

The centripetal force is provided by the tension in the string such that :

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

[tex]v=\sqrt{\dfrac{T_{max}r}{m}}[/tex]

[tex]v=\sqrt{\dfrac{98\times 2}{0.5}}[/tex]

v = 19.79 m/s

or

v = 20 m/s

So, the maximum speed that the mass can attain under these conditions without the string breaking is 20 m/s. Hence, this is the required solution.

Two points are given in polar coordinates,  the distance between the two points is approximately 3.12m.

You can use the polar-to-cartesian conversion formula to convert each point to Cartesian coordinates (x, y), then apply the distance formula in

Cartesian coordinates to determine the separation between two points supplied in polar coordinates.

The polar-to-cartesian conversion formulas are:

[tex]\[ x = r \cdot \cos(\theta) \][/tex]

[tex]\[ y = r \cdot \sin(\theta) \][/tex]

Given the points:

[tex]\( (r_1, \theta_1) = (2.60 \, \text{m}, 50.0^\circ) \)[/tex]

[tex]\( (r_2, \theta_2) = (3.60 \, \text{m}, -46.0^\circ) \)[/tex]

Converting these points to Cartesian coordinates:

For the first point:

[tex]\[ x_1 = 2.60 \, \text{m} \cdot \cos(50.0^\circ) \approx 1.66 \, \text{m} \][/tex]

[tex]\[ y_1 = 2.60 \, \text{m} \cdot \sin(50.0^\circ)\\\\ \approx 1.98 \, \text{m} \][/tex]

For the second point:

[tex]\[ x_2 = 3.60 \, \text{m} \cdot \cos(-46.0^\circ)\\\\ \approx 2.53 \, \text{m} \][/tex]

[tex]\[ y_2 = 3.60 \, \text{m} \cdot \sin(-46.0^\circ)\\\\ \approx -2.57 \, \text{m} \][/tex]

Now, you can use the distance formula in Cartesian coordinates to find the distance between the two points:

[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Plug in the values and calculate:

[tex]\[ \text{distance} = \sqrt{(2.53 \, \text{m} - 1.66 \, \text{m})^2 + (-2.57 \, \text{m} - 1.98 \, \text{m})^2}\\\\ \approx 3.12 \, \text{m} \][/tex]

Thus, the distance between the two points is approximately 3.12 m.

For more details regarding Cartesian coordinates, visit:

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Answer:

option B

Explanation:

given,

seed of airplane = 70 m/s

height = 300 m

we know,

[tex]s = ut + \dfrac{1}{2}at^2[/tex]

[tex]300 = 0 + \dfrac{1}{2} \times 9.81\times t^2[/tex]

t = 7.82 s                                  

now, the range of the crate

 R = V × t

     = 70 × 7.82                      

     = 547.44 ≅ 548 m                          

hence, the correct answer is option B

Answer:

Charge on A is [tex]q=0.7820\times 10^{-5}C[/tex]

Charge on B is [tex]2q=2\times 0.7820\times 10^{-5}C=1.5640\times 10^{-5}C[/tex]  

Explanation:

We have given one charge is twice of other charge

Let [tex]q_1=q[/tex], then [tex]q_2=2q[/tex]

Distance between two charges = 16 cm = 0.16 m

Force F = 43 N

According to coulombs law force between tow charges is given by

[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex], here K is constant which value is [tex]9\times 10^9[/tex]

So [tex]43=\frac{9\times 10^92q^2}{0.16^2}[/tex]

[tex]q^2=0.0611\times 10^{-9}[/tex]

[tex]q^2=0.611\times 10^{-10}[/tex]

[tex]q=0.7820\times 10^{-5}C[/tex] so charge on A is [tex]q=0.7820\times 10^{-5}C[/tex]

And charge on B is [tex]2q=2\times 0.7820\times 10^{-5}C=1.5640\times 10^{-5}C[/tex]

The magnitudes of the charges are:

Charge A: [tex]\underline {1.10 \times 10^{-5} \text{C}}[/tex].

Charge B: [tex]\underline {5.52 \times 10^{-6} \text{C}}[/tex]

Use Coulomb's Law, which states:

[tex]F = k \frac{ |q_1 \, q_2| }{ r^2 }[/tex]

where:

Given:

Let's substitute these values into Coulomb's Law:

[tex]43.0 \; \text{N} = (8.99 \times 10^9 \; \text{Nm}^2/\text{C}^2) \frac{ |2q \cdot q| }{ (0.16 \; \text{m})^2 }[/tex]

Simplify the equation:

[tex]43.0 \; \text{N} = (8.99 \times 10^9) \frac{ 2q^2 }{ 0.0256 } \text{Nm}^2/\text{C}^2[/tex]

[tex]43.0 \; \text{N} = ( 7.03 \times 10^{11} ) 2q^2[/tex]

Solving for [tex]q^2[/tex]:

[tex]43.0 = 1.406 \times 10^{12} q^2[/tex]

[tex]q^2 = \frac{ 43.0 }{ 1.406 \times 10^{12} }[/tex]

[tex]q = \sqrt{ \frac{ 43.0 }{ 1.406 \times 10^{12} } }[/tex]

[tex]q \approx 5.52 \times 10^{-6} \text{C}[/tex]

This gives us the magnitude of charge B [tex]q_2[/tex].

Since [tex]q_1 = 2q_2[/tex]:

Charge A: [tex]q_1 = 2 \times 5.52 \times 10^{-6} \text{C} = 1.10 \times 10^{-5} \text{C}[/tex]

Charge B: [tex]q_2 = 5.52 \times 10^{-6} \text{C}[/tex]

Answer:

F = -10800 N

Explanation:

Given that,

Charge 1, [tex]q_1=-1.5\ C[/tex]

Charge 2, [tex]q_2=0.8\ C[/tex]

Distance between the charges, [tex]d=1\ km=10^3\ m[/tex]

We need to find the electric force acting between two point charges. Mathematically, it is given by :

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

[tex]F=-9\times 10^9\times \dfrac{1.5\times 0.8}{(10^3)^2}[/tex]

F = -10800 N

So, the magnitude of electric force between two point charges is 10800 N. Hence, this is the required solution.

Answer:

106.52 minutes

Explanation:

Given:

Initial distance between Carl and Isaac = 175 miles

speed of Isaac = 65 mph

Speed of Carl = 50 mph

Now, the Carl starts after 35 minutes, but the Isaac has already started so the distance covered by Isaac in 35 minutes will be

= [tex]\frac{35}{60}\times65[/tex]

= 37.91 miles

Therefore,

the distance left between Isaac and Carl = 175 - 37.91 = 137.09 miles

Since Isaac and Carl are moving towards each other,

therefore the relative speed between the both = 65 + 50 = 115 mph

Hence, the time taken to meet = [tex]\frac{137.09}{115}[/tex]

or

The time taken to meet = 1.192 hours

or

The time taken = 1.192 × 60 = 71.52 minutes

Therefore the total time Isaac have been travelling = 71.52 + 35

= 106.52 minutes

Answer:

they meet at distance 25 feet

Explanation:

given data

acceleration of car  = 8 ft/s²

truck speed = 10 ft/s

car initial speed u = 0

truck acceleration = 0

to find out

How far from the starting point will car overtake the truck

solution

we apply here equation of motion

s = ut + 0.5 ×a×t²   .............1

here s is distance and a is acceleration and t is time u is initial speed

so truck distance

s = 10t + 0.5 ×0×t²

s = 10 t   ...............2

and car distance

s = 0+ 0.5 ×8×t²  

s = 4×t²     ..........................3

so from equation 2 and 3

10 t = 4×t²

t = 2.5 s

so both meet at distance

s = 10 (t)

s = 10 ( 2.5 ) = 25 ft

so they meet at distance 25 feet

Answer:

Least velocity.

Explanation:

According to the Bernauli's equation

[tex]p^{2}+\frac{1}{2}\rho v^{2}+\rho gh= constant[/tex]

Here, v is the velocity, m is the mass, h is the height, P is the pressure, [tex]\rho[/tex] is the density

Now according to question.

[tex]P_{1}^{2}+\frac{1}{2}\rho v_{1} ^{2}+\rho gh_{1} =P_{2}^{2}+\frac{1}{2}\rho v_{2} ^{2}+\rho gh_{2}[/tex]

Here airplane height is same means [tex]h_{1}=h_{2}[/tex]  then the required equation will become.

[tex]P_{1}^{2}+\frac{1}{2}\rho v_{1} ^{2}=P_{2}^{2}+\frac{1}{2}\rho v_{2} ^{2}[/tex]

Therefore,

[tex]P_{1}-P_{2}=\frac{1}{2}\rho (v_{2} ^{2}-v_{1} ^{2})[/tex]

Therefore according to the situation [tex]P_{1}>P_{2}[/tex]

This will give the velocity relation [tex]v_{2} >v_{1}[/tex]

Therefore, airplane can fly with least velocity.

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at   ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s  ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so  he travel 36 m to stop

Answer:

value of the acceleration of gravity on the planet is 5.00 m/s²

Explanation:

The problem is similar to a free fall exercise, with another gravity value, the expression they give us is the following:

       y-yo = ½ gₐ t²       (1)

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

  Let's continue our analysis, as we have a linear equation, write the equation of the line.

        y1 = m x1 + b       (2)

where  “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”.  To better see the relationship we rewrite equation 1 with this form

        t² = 2 /gₐ  (y-yo)

With the two expressions written in the same way, let's relate the terms one by one

        y1 = t²

        x1 = (y-yo)

        m = 2/gap

        b= 0

We substitute and calculate

        m = 2/gp

        gₐ = 2/m

        gₐ = 2/ 0.400

        gₐ = 5.00 m / s²

This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant