A car is driven east for a distance of 47 km, then north for 23 km, and then in a direction 32° east of north for 27 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.

Answer:

(a) 13.6 eV

(b) 10.2 V

Explanation:

a) Ionization potential energy is defined as the minimum energy required to excite a neutral atom to its ionized state i.e basically  the minimum energy required to excite an electron from n=1 to infinity.

Energy of a level, n, in Hydrogen atom is, [tex]E_{n}=-\frac{13.6}{n^{2} }[/tex]

Now ionization potential can be calculated as

[tex]E_{\infty}- E_{0}[/tex]

Substitute all the value of energy and n in above equation.

[tex]=-\frac{13.6}{\infty^{2}}-(-\frac{13.6}{1^{2}})\\=13.6eV[/tex]

Therefore, the ionization potential is 13.6 eV.

b) This is the energy required to excite a atom from ground state to its excited state. When electrons jumps from ground state level(n=1) to 1st excited state(n=2) the corresponding energy is called 1st excitation potential energy and corresponding potential is called 1st excitation potential.

So, 1st excitation energy = E(n 2)- E(n = 1)

[tex]=-\frac{13.6}{2^{2}}-(-\frac{13.6}{1^{2}})\\=-3.4eV - (-13.6eV) \\=10.2eV[/tex]

Now we can find that 1st excitation energy is 10.2 eV which gives,

[tex]eV'=10.2eV\\V'=10.2V[/tex]

Therefore, the 1st excitation potential is 10.2V.

Answer:

460 Hz

Explanation:

the given resonating frequency of the device

f₁ =  920 Hz and f₂ =  1380 Hz

fundamental frequency of the device is

f₀ = n₂ - n₁

   = 1380 - 920

   = 460 Hz

expression of frequency of organic pipe open at both ends

[tex]f_0=n\dfrac{\nu}{2l}[/tex]

at n = 1

[tex]f_0=\dfrac{\nu}{2l} = 460 Hz[/tex]

the frequency ratios of the closed pipe

[tex]f_0:f_1:f_2: ...... =[1:2:3:.........]f_0[/tex]

                         =[1:2:3:.........]460 Hz

                         = 460 Hz : 920 Hz : 1380 Hz

so, the lowest frequency for the pipe open at both end is 460 Hz

Answer:

[tex]F = 1361.1 N[/tex]

Explanation:

As we know that a charge is split into two parts

so we have

[tex]q_1 + q_2 = 7\mu C[/tex]

now we have

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we know

[tex]F = \frac{kq_1(7\mu C - q_1)}{r^2}[/tex]

here we have

r = 9 mm

now to obtain the maximum value of the force between two charges

[tex]\frac{dF}{dq_1} = 0[/tex]

so we have

[tex]7 \mu C - q_1 - q_1 = 0[/tex]

so we have

[tex]q_1 = q_2 = 3.5 \mu C[/tex]

so the maximum force is given as

[tex]F = \frac{(9\times 10^9)(3.5 \mu C)(3.5 \mu C)}{0.009^2}[/tex]

[tex]F = 1361.1 N[/tex]

Answer:

7.15 m/s

Explanation:

We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.

The truck moves at constant speed, we can use the equation for position under constant speed:

Xt = X0 + v*t

The car is accelerating with constant acceleration, we can use this equation

Xc = X0 + V0*t + 1/2*a*t^2

We know that both vehicles will meet again at x = 578

Replacing this in the equation of the truck:

578 = 24 * t

We get the time when the car passes the truck

t = 578 / 24 = 24.08 s

Before replacing the values on the car equation, we rearrange it:

Xc = X0 + V0*t + 1/2*a*t^2

V0*t = Xc - 1/2*a*t^2

V0 = (Xc - 1/2*a*t^2)/t

Now we replace

V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s

An equipotential line is like the contour line of a map that had lines of equal altitude. In this case the "altitude"is the electrical potential or voltage. Equipotential lines are always perpendicular to the electric field. In three dimensions these lines form equipotential surfaces. The movement along an equipotential surface does not perform work, because that movement is always perpendicular to the electric field.

Answer:

(a ) vf= 188.12m/s  : Final speed at 8.75 s

(b) d= 823.04 m   : Distance the rocket sled traveled

Explanation:

Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:

d =vi*t+1/2a*t² Formula (1)

vf= vi+at            Formula(2)

Where:

vi: initial speed =0

a: acceleration=21.5 m/s²

t: time=8.75 s

vf: final speed in m/s

d:displacement in meters(m)

Calculation of displacement (d) and final speed (vf)

We replace data in formulas (1) and (2):

d= 0+1/2*21.5*8.75²

d= 823.04 m

vf= 0+21.5*8.75

vf= 188.12m/s

Answer:

[tex]C_{eq}=1.97\ \mu F[/tex]

Explanation:

Given that,

Capacitance 1, [tex]C_1=0.5\ \mu F[/tex]

Capacitance 2, [tex]C_2=11\ \mu F[/tex]

Capacitance 3, [tex]C_3=1.5\ \mu F[/tex]

C₁ and C₂ are connected in series. Their equivalent is given by :

[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]

[tex]\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}[/tex]

[tex]C'=0.47\ \mu F[/tex]

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

[tex]C_{eq}=C'+C_3[/tex]

[tex]C_{eq}=0.47+1.5[/tex]

[tex]C_{eq}=1.97\ \mu F[/tex]

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

Answer:[tex]0.0686 m^3[/tex]

Explanation:

Given

No of moles=2

[tex]T=25^{\circ}\approx 298 K[/tex]

[tex]P_1=210 KPa[/tex]

[tex]T_2=48 ^{\circ}\approx 321 k[/tex]

[tex]P_2=77.7 KPa[/tex]

PV=nRT

Substitute

[tex]77.7\times V_2=2\times 8.314\times 321[/tex]

[tex]V_2=0.0686 m^3[/tex]

Answer:

Its dissipated by the brake.

Explanation:

Traditional brakes use friction to stop the wheels (or axis). This friction its dissipated in the form of heat. There are others mechanism to brake that don't dissipated the energy, they stored it. In electric cars (or hybrids), there are regenerative brakes, that store the kinetic energy as electrical energy.

As the driver slams the brakes, the kinetic energy of the car gets converted into other forms of energy until the car comes to a stop, at which point it becomes zero.

When the driver of a car slams on her brakes to avoid a collision with a deer crossing the highway, the kinetic energy of the car decreases until the car comes to rest. This is due to the principle of energy conservation. Initially, when the car is moving, it has kinetic energy. However, when the brakes are applied, the kinetic energy gets converted into other forms of energy such as heat energy (due to friction between the brakes and the wheels) and potential energy (if the car is moving uphill). The kinetic energy keeps decreasing until the car comes to a complete stop. At that point, the kinetic energy of the car is zero because kinetic energy is associated with motion and the car is no longer moving.

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Answer with Explanation:

The general wave equation is given by

[tex]\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=\frac{1}{c^2}\frac{\partial ^2u}{\partial t^2}[/tex]

where

'c' is the velocity of the wave

Comparing with the given equation

[tex]\frac{\partial ^2h}{\partial x^2}+\frac{\partial ^2h}{\partial y^2}=\frac{1}{gd}\frac{\partial ^2h}{\partial t^2}[/tex]

We can see that

[tex]c^2=gd\\\\\therefore c=\sqrt{gd}[/tex]

Thus the velocity of  wave is given by [tex]v=\sqrt{gd}[/tex]

Answer:

A) 29.5m/s

B) 25.9m/s

C) 47.9 m/s

Explanation:

This is a relative velocity problem, which means that the velocity perception will vary from each observer at a different reference point.

We can say that the velocity of the train respect the ground is 28m/s on east direction If I walk 1.5m/s respect the train, the velocity of the person respect to the ground is the sum of both velocities:

Vgp=Vgt+Vtp

where:

Vgp=velocity of the person respect the ground

Vgt=Velocity of the train respect the ground

Vtp=Velocity of the person respect the train

1) Vgp=28m/s+1.5m/s=29.5m/s

2) here the velocity of the person respect the train is negative because it is going backward

Vgp=28m/s-2.1m/s=25.9m/s

3) the velocity of the train respect with your friend will be the sum of both velocities(Vft).

Vfp=Vft+Vtp

Vfp=velocity of the person respect the friend

Vft=Velocity of the train respect the friend

Vtp=Velocity of the person respect the friend

Vfp=(22m/s+28m/s)+(-2.1m/s)=47.9 m/s

Answer:

speed when the block had slid 3.40 m is 2.68 m/s

Explanation:

given data

distance = 6.80 m

speed = 3.80 m/s

to find out

speed when the block had slid 3.40 m

solution

we will apply here equation of motion that is

v²-u² = 2×a×s   ..............1

here s is distance, a is acceleration and v is speed and u is initial speed that is 0

so put here all value in equation 1 to get a

v²-u² = 2×a×s

3.80²-0 = 2×a×6.80

a = 1.06 m/s²

so

speed when distance 3.40 m

from equation 1 put value

v²-u² = 2×a×s

v²-0 = 2×1.06×3.40

v² = 7.208

v = 2.68

so speed when the block had slid 3.40 m is 2.68 m/s

Answer:

1) Distance traveled equals 23.1 meters.

2) Final velocity equals 21.658 m/s.

Explanation:

The problem can be solved using second equation of kinematics as

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

where

s is the distance covered

u is the initial speed of the ball

a is the acceleration the ball is under

t is time of travel

Applying the given values in the above equation we get

[tex]s=4.0\times 1.8+\frac{1}{2}\times 9.81\times 1.8^{2}\\\\s=23.1meters[/tex]

Part 2)

The velocity after 't' time can be obtained using first equation of kinematics.

[tex]v=u+at[/tex]

Applying the given values we get

[tex]v=4+9.81\times 1.80\\\\\therefore v=21.658m/s[/tex]

Final answer:

The ball travels 23.07 meters in 1.80 seconds, and the velocity of the ball at that time is 21.66 m/s downward.

Explanation:

The student wants to know how far a ball projected vertically downward at a speed of 4.00 m/s will travel in 1.80 s and the ball's velocity at that time. To solve these questions, we use the kinematic equations that govern linear motion with constant acceleration due to gravity, ignoring air resistance.

The distance traveled by the ball can be found using the equation:

s = ut +  [tex]rac{1}{2}at^2[/tex]

Where:

s is the distance traveled,

u is the initial velocity (4.00 m/s),

t is the time (1.80 s),

a is the acceleration due to gravity (9.81 m/s^2).

Plugging in the values, we have:

s = 4.00 m/s  imes 1.80 s + 0.5  imes 9.81 [tex]m/s^2[/tex] imes [tex](1.80 s)^2[/tex]

s = 7.20 m + 15.87 m

s = 23.07 m

The ball travels 23.07 meters in 1.80 seconds.

The velocity of the ball after 1.80 seconds can be calculated using the equation:

v = u + at

Where:

v is the final velocity,

u is the initial velocity,

a is the acceleration,

t is the time.

So, v = 4.00 m/s + 9.81 [tex]m/s^2[/tex] imes 1.80 s

v = 4.00 m/s + 17.66 m/s

v = 21.66 m/s

The velocity of the ball after 1.80 seconds is 21.66 m/s downward.

Answer:0.835 s

Explanation:

Given

Red ball initial velocity([tex]u_r[/tex])=1.1 m/s

height of building(h)=28 m

after 0.5 sec blue ball is thrown with a velocity([tex]u_b[/tex])=24.3 m/s

Height of blue after 2 sec red ball is thrown

i.e. height of blue ball at t=1.5 sec after blue ball is thrown upward

[tex]h=24.3\times 1.5-\frac{9.81\times 1.5^2}{2}=25.414 m[/tex]

therefore blue ball is at height of 25.414+0.9=26.314 from ground

moment after the two ball is at same height

for red ball

[tex]14=1.1\times \left ( t+0.5\right )+\frac{9.81\times \left ( t+0.5\right )^2}{2}-----1[/tex]

for blue ball

[tex]13.1=24.3\times t-\frac{9.81\times t^2}{2}-----2[/tex]

add 1 & 2

we get

[tex]27.1=1.1t+0.55+24.3t+\frac{g\left ( t+0.25\right )}{2}[/tex]

27.1=25.4t+0.55+4.905t+1.226

t=0.835 s

The magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]

To find the tension on the ends of the clothesline, we can use the principle of equilibrium. When the mass is at rest in the middle of the clothesline, the tension in the clothesline must balance the gravitational force acting on the mass.

Let's denote:

- [tex]\( T \)[/tex] as the tension in the clothesline.

- [tex]\( m \)[/tex] as the mass tied to the clothesline.

- [tex]\( g \)[/tex] as the gravitational acceleration.

The gravitational force acting on the mass is given by [tex]\( F_{\text{gravity}} = m \cdot g \).[/tex]

When the mass is at rest in the middle of the clothesline, the horizontal components of the tensions on either side cancel each other out, leaving only the vertical components to support the weight.

Since the clothesline sags a distance of 3 meters in the middle, each side of the clothesline forms a right triangle with the sagging distance as one leg and half of the distance between the poles as the other leg. Therefore, each side of the clothesline forms a right triangle with legs of 8 meters and 3 meters.

Using the Pythagorean theorem, we can find the length of the clothesline (the hypotenuse of each triangle), which is the same as the length of each segment between the midpoint and the poles:

[tex]\[ L = \sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} \]\[ L = \sqrt{64 + 9} \]\[ L = \sqrt{73} \][/tex]

Now, we can set up an equation for the vertical equilibrium:

[tex]\[ 2T \sin(\theta) = F_{\text{gravity}} \][/tex]

Where [tex]\( \theta \)[/tex] is the angle the clothesline makes with the horizontal.

We can find [tex]\( \sin(\theta) \)[/tex] using trigonometry:

[tex]\[ \sin(\theta) = \frac{3}{L} \][/tex]

Now, substitute this expression for [tex]\( \sin(\theta) \)[/tex] into the equation for vertical equilibrium:

[tex]\[ 2T \left(\frac{3}{L}\right) = m \cdot g \]\[ 2T \left(\frac{3}{\sqrt{73}}\right) = 3 \times 9.8 \]\[ T = \frac{3 \times 9.8 \times \sqrt{73}}{2 \times 3} \]\[ T = \frac{9.8 \times \sqrt{73}}{2} \][/tex]

Now, we can calculate the tension:

[tex]\[ T \approx \frac{9.8 \times \sqrt{73}}{2} \]\[ T \approx \frac{9.8 \times 8.544}{2} \]\[ T \approx \frac{83.9712}{2} \]\[ T \approx 41.9856 \][/tex]

So, the magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]

The magnitude of the tension at the ends of the clothesline is approximately [tex]\[T = 41.87 \, \text{N}\][/tex]

To solve for the tension in the clothesline, we first consider the forces and the geometry involved.

Given:

Distance between poles, [tex]\( L = 16 \) meters[/tex]

Sagging distance,[tex]\( h = 3 \) meters[/tex]

Mass, [tex]\( m = 3 \) kilograms[/tex]

Gravitational acceleration, [tex]\( g = 9.8 \) m/s\(^2\)[/tex]

Determine the Weight of the Mass

The weight [tex]\( W \)[/tex] of the mass is given by:

[tex]\[W = mg = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{N}\][/tex]

Analyzing the Geometry

[tex]\[\text{Half the distance between poles} = \frac{L}{2} = \frac{16}{2} = 8 \, \text{meters}\][/tex]

[tex]\[\text{Sagging distance} = h = 3 \, \text{meters}\][/tex]

Calculate the Length of the Hypotenuse

[tex]\[\sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} = \sqrt{64 + 9} = \sqrt{73} \, \text{meters}\][/tex]

Analyze Forces in Vertical and Horizontal Directions

[tex]\[2T_v = W \quad = \quad T_v = \frac{W}{2} = \frac{29.4 \, \text{N}}{2} = 14.7 \, \text{N}\][/tex]

The vertical component of the tension [tex]\( T \)[/tex] in each half of the clothesline is:

[tex]\[T_v = T \sin \theta\][/tex]

where \( \theta \) is the angle the clothesline makes with the horizontal. From the triangle, we know:

[tex]\[\sin \theta = \frac{h}{\sqrt{(L/2)^2 + h^2}} = \frac{3}{\sqrt{73}}\][/tex]

[tex]\[T \sin \theta = 14.7 \, \text{N}\][/tex]

Solve for the Tension[tex]\( T \)[/tex]

[tex]\[T \sin \theta = 14.7 \, \text{N} \quad = \quad T \cdot \frac{3}{\sqrt{73}} = 14.7 \, \text{N}\][/tex]

[tex]\[T = \frac{14.7 \times \sqrt{73}}{3}\][/tex]

[tex]\[T = 14.7 \times \frac{\sqrt{73}}{3}\][/tex]

[tex]\[T = 14.7 \times \frac{8.544}{3}\][/tex]

[tex]\[T = 14.7 \times 2.848 = 41.87 \, \text{N}\][/tex]

Answer:

Explanation:

The weight w its:

[tex]w \ = \ m \ a[/tex]

where m is the mass and a is the acceleration.

Local acceleration of gravity = [tex]9.779 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.779 \frac{m}{s^2}[/tex]

[tex]w = \ 782.32 \ N [/tex]

Local acceleration of gravity = [tex]9.796 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.796 \frac{m}{s^2}[/tex]

[tex]w = \ 783.68 \ N [/tex]

Local acceleration of gravity = [tex]9.798 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.798 \frac{m}{s^2}[/tex]

[tex]w = \ 783.84 \ N [/tex]

Local acceleration of gravity = [tex]9.803 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.803 \frac{m}{s^2}[/tex]

[tex]w = \ 784.24 \ N [/tex]

Local acceleration of gravity = [tex]9.815 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.815 \frac{m}{s^2}[/tex]

[tex]w = \ 785.2 \ N [/tex]

Answer

mass of a person is 80 kg

to calculate weight of the person = ?

a) Mexico City, Mexico

acceleration due to gravity in mexico = 9.779 m/s²

   weight = 80× 9.779 = 782.32 N

b) Cape Town, South Africa

acceleration due to gravity in cape town = 9.796 m/s²

   weight = 80× 9.796 = 783.68 N

c) Tokyo, Japan

acceleration due to gravity in Tokyo = 9.798 m/s²

   weight = 80× 9.798 = 783.84 N

d) Chicago, IL

acceleration due to gravity in Chicago = 9.80 m/s²

   weight = 80× 9.80 = 784 N

e) Copenhagen, Denmark

acceleration due to gravity in Copenhagen = 9.81 m/s²

   weight = 80× 9.81 = 784.8 N

Answer:

The equivalent capacitance will be [tex]15\mu F[/tex]  

Explanation:

We have given two capacitance [tex]C_=10\mu F\ and\ C_2=20\mu F[/tex]

They are connected in parallel

So equivalent capacitance [tex]C=C_1+C_2=10+20=30\mu F[/tex]

This equivalent capacitance is now connected in series with [tex]30\mu F[/tex]

In series combination of capacitors the equivalent capacitance is given by [tex]\frac{1}{C}=\frac{1}{30}+\frac{1}{30}[/tex]

[tex]C=\frac{30}{2}=15\mu F[/tex]

So the equivalent capacitance will be [tex]15\mu F[/tex]  

The equivalent capacitance of the arrangement where capacitors of 10 uF and 20 uF are connected in parallel, and in series with a 30 uF capacitor, is 15 uF.

To determine the equivalent capacitance of a combination in which capacitors are connected both in parallel and in series, one must consider the rules for each type of connection. For capacitors connected in parallel, the equivalent capacitance is simply the sum of their individual capacitances. When capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. In your case, the 10 uF and 20 uF capacitors are connected in parallel, so their equivalent capacitance is 10 uF + 20 uF = 30 uF. This combined capacitance is then in series with the 30 uF capacitor.

The next step is to calculate the equivalent capacitance for the series connection using the formula:

1/Cseries = 1/C1 + 1/C2

Where C1 is the equivalent parallel capacitance (30 uF) and C2 is the capacitance of the third capacitor (30 uF). Substituting the values:

1/Cseries = 1/30 uF + 1/30 uF = 1/15 uF

So, Cseries = 15 uF. This is the total equivalent capacitance of the combination of capacitors.

Answer:

The answer is: 0.00563 mg = 5.63 μg

Explanation:

Significant figures of a number are the figures or digits that carry meaning and contributes to the precision of the given number.

The given quantity 0.00563 mg, has three significant figures as all the leading zeros are not significant.

Now to represent 0.00563 mg between 0.1 and 1000, the given unit milligram (mg) is converted to microgram (μg).

As, 1 mg = 1000 mg

Therefore, 0.00563 mg = 5.63 μg, has three significant figures as non-zeros figures are significant.

Therefore, 5.63 μg lies between between 0.1 and 1000 and has significant figures.

The speed of the boat with respect to the shore is 1.91 m/s.

From the information given, we have that;

A boat's speed in still water is 1.60 m/s

The boat is to travel directly across a river whose current has speed 1.05 m/s

We can see that the movement is in both horizontal and vertical directions.

Using the Pythagorean theorem, let use determine the resultant speed of the boat with respect to the shore, we have that;

Resultant speed² = √((boat's speed)² + (current's speed)²)

Substitute the value as given in the information, we have;

= (1.60)² + (1.05 )²)

Find the value of the squares, we get;

= (2.56 + 1.1025 )

Find the square root of both sides, we have;

=  √3.6625

Find the square root of the value, we have;

= 1.91 m/s.

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Answer:

Work done,[tex]w=5.12\times10^{6}\ \rm J[/tex]

change in internal Energy ,[tex]\Delta U=5.12\times10^6\ \rm J[/tex]

Explanation:

Given:

Since It is given that the process is adiabatic process it means that there is no exchange of heat between the system and surroundings

[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}\\\\286\times V_i^{\gamma -1}=T_f \left( \dfrac{V_i}{5} \right )^{\gamma -1}\\T_f=840.76\ \rm K[/tex]

Let n be the number of moles of Helium given by

[tex]n=\dfrac{m}{M}\\n=\dfrac{3\times10^3}{4}\\n=0.75\times10^3[/tex]

Work done in Adiabatic process

Let W be the work done

[tex]W=\dfrac{nR(T_1-T_2)}{\gamma-1}\\W=\dfrac{0.75\times10^3\times8.314(286-840.76)}{1.67-1}\\W=-5.12\times 10^6\ \rm J[/tex]

The Internal Energy change in any Process is given by

Let [tex]\Delta U[/tex] be the change in internal Energy

[tex]\Delta U=nC_p\Delta T\\\Delta U=0.75\times10^3\times1.5R\times(840.76-286)\\\Delta U=5.12\times10^6\ \rm J[/tex]

Answer:

22.02 lb

Explanation:

The weight of astronaut Earth (w) = 588 N

We know that,

[tex]4.45N = 1lb[/tex]

Thus,

[tex]588N = \frac{1}{4.45}\times 588lb[/tex]

588N = 132.13 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

[tex]g'=\frac{g}{6}[/tex]

We know that weight of an object on Earth is,

[tex]w = m\times g[/tex]

[tex]m = \frac{w}{g}[/tex]

Similarly, weight on Moon will be

[tex]w' = m\times g'[/tex]

[tex]w' = \frac{w}{g}\times\frac{g}{6}[/tex]

[tex]w' = \frac{132.13}{6}[/tex]

[tex]w' = 22.02[/tex]

Thus the astronaut will weigh 22.02 lb on Moon.

Answer:

a) 2034 J

b) -1471 J

c) -509 J

Explanation:

The trunk has a mass of 50 kg, so its weight is

f = m * a

f = 50 * 9.81 = 490 N

If the incline is of 30 degrees, the force tangential to the incline is:

ft = f * sin(a)

ft = 490 * sin(30) = 245 N

And the normal force is:

fn = f * cos(a)

fn = 490 * cos(30) = 424 N

The frictional force is:

ff  = μ * fn

ff = 0.2 * 424 = 84.8 N

To push the trunk up one must apply a force slightly greater than the opposing forces, the opposing are the tangential component of the weight and the friction force

fp = fn + ff

fp = 245 + 84.8 = 339 N

The work of the applied force is:

L = f * d

Lp = fp * d

Lp = 339 * 6 = 2034 J

The work of the weight is done by the tangential component:

Lw = -245 * 6 = -1471 J (it is negative because the weight was opposed to the direction of movement)

The work of the friction force is

Lf = -84.8 * 6 = -509 J

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

[tex]|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82[/tex]

To calculate the final velocity of someone running with constant acceleration, we first find the acceleration using the distance and time, and then apply it to determine the final velocity. The final velocity after 10 seconds is 8 m/s.

To find the final velocity of the person running with constant acceleration, we use the kinematic equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. The person starts from rest, so u = 0. We need to find a first, and we can do so using the equation s = ut + (1/2)at², where s is the distance traveled.

Insert the given values: 40m = 0 + (1/2)a(10s)², resulting in a = 0.8 m/s². Now apply the acceleration to the initial equation: v = 0 + 0.8 m/s² * 10s, which gives us v = 8 m/s. Therefore, the person's final velocity after 10 seconds is 8 m/s.

Answer:

The final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

Explanation:

A ball is dropped from the height h. As the ball is dropped means there is no initial velocity given to the ball.

GIVEN:

The distance traveled by the ball is equal to h.

Initial velocity of the ball is zero.

Concept:

Two type of force comes into play in dropping of ball.

1. Gravitational force

Due to gravitational force ball forced to fall downward towards the surface.

2. Air resistance

Air develops drag force on the ball in opposite direction of the gravitational force.

Assumption:

There is no air resistance in falling of ball. So, there will be no drag force on the ball.

Calculation:

It is given that a ball is dropped from a building of height h.

The Newton’s equations of motion are as follows:

[tex]V^{2}-u^{2}=2as[/tex]

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

[tex]v=u+at[/tex]

STEP 1

Newton’s equation of motion is expressed as follows:

[tex]V^{2}-u^{2}=2as[/tex]

Here, V is final velocity of the ball just before the ball hits the ground, a is acceleration due to gravity, u is the initial velocity of the ball and s is the distance traveled by the ball.

Here, initial velocity is zero. So, u = 0 m/s.

Vertical distance traveled by the ball is h. So, s= h.

Only gravitational force is responsible for motion as well as for acceleration. Thus, the acceleration due to gravity acts on the ball. Acceleration due to gravity is denoted by small g.

Acceleration due to gravity is constant for different planate. The acceleration due to gravity for the earth is 9.8 m/s². Here, no planet is defined. So the acceleration due to gravity is a = g.

Substitute 0 for m/s, g for a and h for s in above equation as follows:

[tex]V^{2}-u^{2}=2as[/tex]

[tex]V^{2}-(0)^{2}=2gh[/tex]

[tex]V^{2}=2gh[/tex]

[tex]v=\sqrt{2gh}[/tex]

Thus, the final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

Answer:

Part a)

[tex]T = 59.5 s[/tex]

Part b)

v = 22.8 m/s

Explanation:

Part a)

Time taken by the taxi to reach the maximum speed is given as

[tex]v_f - v_i = at[/tex]

[tex]27 - 0 = 2(t)[/tex]

[tex]t = 13.5 s[/tex]

now it moves with constant speed for next 41 s and then finally comes to rest in next 5 s

so total time for which it will move is given as

[tex]T = 13.5 s + 41 s + 5 s[/tex]

[tex]T = 59.5 s[/tex]

Part b)

Distance covered by the taxi while it accelerate to its maximum speed is given as

[tex]d_1 = \frac{v_f + v_i}{2} t[/tex]

[tex]d_1 = \frac{27 m/s + 0}{2} (13.5)[/tex]

[tex]d_1 = 182.25[/tex]

Now it moves for constant speed for next 41 s so distance moved is given as

[tex]d_2 = (27)(41)[/tex]

[tex]d_2 = 1107 m[/tex]

Finally it comes to rest in next 5 s so distance moved in next 5 s

[tex]d_3 = \frac{v_f + v_i}{2} t[/tex]

[tex]d_3 = \frac{27 + 0}{2}(5)[/tex]

[tex]d_3 = 67.5 m[/tex]

so here we have total distance moved by it is given as

[tex]d = 182.25 + 1107 + 67.5 [/tex]

[tex]d = 1356.75 m[/tex]

average speed is given as

[tex]v = \frac{1356.75}{59.5}[/tex]

[tex]v = 22.8 m/s[/tex]

Final answer:

The taxi is in motion for 59.5 seconds and its average velocity is 19.38 m/s.

Explanation:

(a) To find the total time the taxi is in motion, we need to calculate the time it takes to accelerate to 27.0 m/s, the time it travels at constant speed, and the time it takes to stop. The time to accelerate is given by the equation:

t = (vf - vi) / a = (27.0 m/s - 0 m/s) / 2.00 m/s² = 13.5 s

The time traveling at constant speed is given as 41.0 s, and the time to stop is given as 5.00 s. Therefore, the total time the taxi is in motion is:

Total time = Time to accelerate + Time at constant speed + Time to stop = 13.5 s + 41.0 s + 5.00 s = 59.5 s

(b) The average velocity is given by the total displacement divided by the total time. The displacement during acceleration is given by:

Displacement = vi × t + 0.5 × a × t² = 0 m/s × 13.5 s + 0.5 × 2.00 m/s² × (13.5 s)² = 182.25 m

The displacement during constant speed is equal to the constant speed multiplied by the time:

Displacement = constant speed × time = 27.0 m/s × 41.0 s = 1,107 m

The displacement during deceleration is given by:

Displacement = vf × t + 0.5 × (-a) × t² = 27.0 m/s × 5.00 s + 0.5 × (-2.00 m/s²) × (5.00 s)² = -137.5 m

Therefore, the total displacement is 182.25 m + 1,107 m - 137.5 m = 1,152.75 m.

The average velocity is then:

Average velocity = Total displacement / Total time = 1,152.75 m / 59.5 s = 19.38 m/s

Answer:

10 years

Explanation:

As you can understand from the question it is given that the planet is already filled to half of its capacity. Also the population doubles in 10 years. To fill up the planet completely the population needs to double only once. To do that only 10 years are required.

As it is mentioned there are no other factors affecting the growth rate, in 10years the planet will be filled to its carrying capacity.

Answer:

Magnitude of resultant force is  1190.314 N

Direction of force is 13.352°

Explanation:

given data:

thrust force = 725 N

Angle = 32.4 degree

Let x is consider as positive direction

Resultant force in x direction is  

Rx = 725 + 513cos32.4 = 1158.14 N

and Resultant force perpendicular to x direction is:

Ry = 513sin32.4 = 274.88 N

Magnitude of resultant force is

[tex]R=\sqrt{R_x^2+R_y^2} = 1190.314N[/tex]

and resultant force direction is  

[tex]\theta=tan^{-1}\frac{R_y}{R_x} = 13.352\degree[/tex]

Answer:

The charge on each plate of the capacitor is [tex]19.38 \mu C[/tex]

with one plate positive and one negative, i.e., [tex]\pm 19.38 \mu C[/tex]

Solution:

According to the question:

Critical value of Electric field, [tex]E_{c} = 3.0\times 10^{6} V/m[/tex]

Area of each plate of capacitor, [tex]A_{p} =73 cm^{2} = 73\times 10^{- 4} m^{2}[/tex]

Now, the amount of charge on the capacitor's plates can be calculated as:

Capacitance, C  = [tex]\frac{epsilon_{o}\times Area}{Distance, D}[/tex]        (1)

Also, Capacitance, C = [tex]\frac{charge, q}{Voltage, V}[/tex]

And

Electric field, E = [tex]\frac{Voltage, V}{D}[/tex]

So, from the above relations, we can write the eqn for charge, q as:

q = [tex]\epsilon_{o}\times E_{c}\times A_{p}[/tex]

q = [tex]8.85\times 10^{- 12}\times 3.0\times 10^{6}\times 73\times 10^{- 4}[/tex]

[tex]q = 19.38 \mu C[/tex]

Answer:

Explanation:

This problem can be solved easily if we represent velocity in the form of vector.

The velocity of 351 was towards easterly direction so

V₁ = 351 i

The velocity of 351 was towards south west making - 48° with east or + ve x direction.

V₂ = 351 Cos 48 i - 351 sin 48 j

V₂ = 234.86 i - 260.84 j

Change in velocity

= V₂ - V₁ = 234.86 i - 260.84 j - 351 i

= -116.14 i - 260.84 j

acceleration

= change in velocity / time

(-116.14 i - 260.84 j )/ 1

= -116.14 i - 260.84 j

magnitude = 285.53 ms⁻²

Direction

Tan θ = 260.84 / 116.14 = 2.246

θ = 66 degree south of west .