A 1500kg space probe is moving at constant velocity of 4200m/sec. In a course change maneuver the rocket engines fire for 5 minutes increasing the probe's velocity to 5000m/sec. What is the probe's change in kinetic energy? A. 3.68x10^6 J. B. 5.52x10^9 J C. 1.1x10^10 J D. not enough info.

Answer:

Explanation:

The speed of hare = .15 x 25 = 3.75 m /s . Let tortoise took t second to complete the race .

Distance traveled by it = .15 t

Distance traveled by hare = .15 t - .35 m

Time taken by hare to complete this distance

=  t - 5 x 60 s

Speed of hare

= Distance / time

( .15t-.35 ) / t - 300 , so

[tex]\frac{.15t-.35}{t-300} = 3.75[/tex]

t = 312.40

= 5 minutes 12.4 seconds

Distance of race

312.4 x speed of tortoise

= 312.4 x .15

= 46.85 m

(a) The length of the race is 91.1 meters.

(b) The race takes 605.0 seconds.

The tortoise moves at a speed of 0.15 m/s while the hare moves at a speed of 25 * 0.15 = 3.75 m/s.

The hare stops for 5 minutes, which is equal to 5 * 60 = 300 seconds.

In the 300 seconds that the hare was taking a nap, the tortoise was able to move a distance of 0.15 * 300 = 45 meters.

When the hare woke up, he started running at his top speed. He was able to catch up to the tortoise and surpass him by 35 cm. This means that the hare was able to run a distance of 45 + 0.35 = 45.35 meters.

Therefore, the length of the race is 45.35 meters.

The total time it took for the race to finish is 300 + 605 = 905 seconds.

So the answer is (a) 91.1 meters (b) 605.0 seconds

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Explanation:

The average speed is the  displacement of a particle divided into a time in which the displacement occurs. Instead, instantaneous speed is the limit of the average speed as the time interval approaches zero.

In physics expressions

[tex]Average speed=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }[/tex]

Instantaneous speed=[tex]lim_{t \to \infty} \frac{x}{t} =\frac{dx}{dt}[/tex]

b) Speed ​​denotes distance traveled divided by time, and instantaneous speed measures how quickly and in which direction does it move

The difference between average speed and instantaneous speed is that average speed is the total distance traveled divided by the total time taken, whereas instantaneous speed is the speed of an object at a particular moment in time. Speed is a scalar quantity which means it only has magnitude and no direction, while velocity is a vector quantity which includes both magnitude and direction.

Average speed is calculated over a stretch of time and it does not consider the direction of the motion. For instance, if you drove for 100 kilometers and it took you 2 hours, your average speed would be 50 kilometers per hour. On the other hand, instantaneous speed is the speed of an object at a specific instant. It can be seen as the speedometer reading in a vehicle at a particular moment. To illustrate, while driving, your speedometer might read 60 km/h at one instant, which would be your instantaneous speed at that time.

Moreover, velocity includes direction as part of its definition and it describes the rate of change of an object's position. Average velocity takes into account the total displacement (change in position) and the total time. If the same 100 kilometers were travelled in a direct line to the north and it took 2 hours, then the average velocity would be 50 km/h north. If, however, the trip involved a return to the starting point (making the total displacement zero), the average velocity would also be zero, despite having a non-zero average speed. Therefore, while they might have the same magnitude under certain circumstances, speed and velocity are not interchangeable terms due to velocity's directional component.

Understanding the distinction between speed and velocity is crucial in physics because it determines how motion is described and analyzed.

Answer:

(a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

Explanation:

Given that,

A bird watcher meanders through the woods, walking 1.46 km due east, 0.123 km due south, and 4.24 km in a direction 52.4° north of west.

(a). We need to calculate the displacement

Using Pythagorean theorem

[tex]D=\sqrt{(OA)^2+(AB)^2}[/tex]

[tex]D=\sqrt{(1.46-4.24\sin52.4)^2+(0.123-4.24\cos52.4)^2}[/tex]

[tex]D=3.11\ km[/tex]

(b). We need to calculate the average velocity

Using formula of average velocity

[tex]v_{avg}=\dfrac{D}{T}[/tex]

Where, D = displacement

T = time

Put the value into the formula

[tex]v_{avg}=\dfrac{3.11}{1.285}[/tex]

[tex]v_{avg}=2.42\ km/hr[/tex]

Hence, (a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

Answer:

The correct answer is option 'a' 'The momentum is always conserved while as the kinetic energy may be conserved'

Explanation:

The conservation of momentum is a basic principle in nature which is always valid in an collision between 'n' number of objects if there are no external forces on the system. It is valid for both the cases weather the collision is head on or glancing or weather the object is elastic or inelastic.

The energy is only conserved in a collision that occurs on a friction less surface and the objects are purely elastic. Since in the given question it is mentioned that only the surface is friction less and no information is provided regarding the nature of the objects weather they are elastic or not hence we cannot conclusively come to any conclusion regarding the conservation of kinetic energy as the objects may be inelastic.

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

1) a = 2.13 m/s²

2) F = 51.12 N

3) T_max = 851.62 N

4)a₁ = 6.08 m/s²

5) a₂ = 12.11 m/s²

This is based on the concept of frictional motion with coefficients of friction.

We are given;

Mass of smaller top crate; m₁ = 24 kg

Mass of larger bottom crate; m₂ = 86 kg

Coefficient of static friction; μ_s = 0.79

Coefficient of kinetic friction; μ_k = 0.62

For the top crate, the sum of forces will be expressed as;

F = m₁a₁  - - - (eq 1)

For the bottom crate, the sum of forces will be expressed as;

T - F = m₂a₂  - - - (eq 2)

Where F is frictional force exerted by small crate on big crate and T is tension.

1) We are told that the rope is pulled with a tension T = 234 N. And that the top crate will not slide. Thus, the acceleration of the small crate will be gotten from a combination of equation 1 and 2 to get the formula;

T = m₂a + m₁a

Making a which is the acceleration the subject gives;

a = T/(m₁+m₂)

Plugging in the relevant values gives;

a = 234/(24 + 86)

a = 2.13 m/s²

2) From eq(1) above, we see that the formula for the frictional force the small crate exerts on the big one is;

F = m₁a₁

Thus, Plugging in the relevant values gives;

F = 24 × 2.12

F = 51.12 N

3) maximum tension that the lower crate can be pulled at before the upper crate begins to slide will occur at the maximum friction. The formula for mx friction is;

(F₂)_max = μ_s•m₂•g

Using the concept of eq(1) We can say that;

F₂ = m₂a

m₂a = μ_s•m₂•g

m₂ will cancel out to give;

a = μ_s • g

From first answer earlier, we saw that;

T = m₂a + m₁a

T = (m₂ + m₁)a

Plugging in the relevant values gives;

T_max = (m₁ + m₂)μ_s*g

T_max = (24 + 86) × 0.79 × 9.8

T_max = 851.62 N

4) To get the acceleration of the upper crate, we will make use of the formula for kinetic friction which is:

F = μ_k*m₁g

From earlier, we saw that; F = m₁a₁

Thus;

m₁a₁ = μ_k*m₁g

m₁ will cancel out to get;

a₁ = μ_k*g

a₁ = 0.62 × 9.8

a₁ = 6.08 m/s²

5) the acceleration of the lower crate as the upper crate slides will be gotten by putting μ_k*m₁*g for m₁a₁ in eq(2) to get;

m₂a₂ = T - (μ_k*m₁*g)

Making a₂ the subject gives us;

a₂ = (T - μ_k•m₁•g)/m₂

Plugging in the relevant values;

a₂ = (1187 - (0.62 × 24 × 9.8))/86

a₂ = 12.11 m/s²

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Answer:

A) x and y components of the electric field  (Ep) at the origin.

Epx = -1620.5 N/C

Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

   Ep= 2291.7 N/C

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)

Equivalence  

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P  (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point  P (x=0, y=0) due to charge q₂. As the charge q₂  is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field  (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

[tex]E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}[/tex]

Answer:

a) [tex]v_{3} =8.43 m/s[/tex]

b) [tex]v_{2}=2.15m/s[/tex]

c) ΔK=[tex]-28.18x10^4J[/tex]

d)ΔK=[tex]-10.33x10^4J[/tex]

Explanation:

From the exercise we know that there is a collision of a sports car and a truck.

So, the sport car is going to be our object number 1 and the truck object number 2.

[tex]m_{1}=1050kg\\v_{1}=-13m/s\\m_{2}=6320kg\\v_{2}=12m/s[/tex]

Since the two vehicles remain locked together after the collision the final mass is:

[tex]m_{3}=7370kg[/tex]

a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle

[tex]p_{1}=p_{2}[/tex]

[tex]m_{1}v_{1}+ m_{2}v_{2}=m_{3}v_{3}[/tex]

[tex]v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3}}=\frac{(1050kg)(-13m/s)+(6320kg)(12m/s)}{7370kg}[/tex]

[tex]v_{3}=8.43m/s[/tex]

b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before

[tex]m_{1}v_{1}+ m_{2}v_{2}=0[/tex]

[tex]v_{2}=\frac{-m_{1}v_{1}}{m_{2} }=\frac{-(1050kg)(-13m/s)}{(6320kg)}=2.15m/s[/tex]

c) To find the change in kinetic energy we need to do the following steps:

ΔK=[tex]k_{2}-k_{1}=\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]

ΔK=[tex]\frac{1}{2}(7370)(8.43)^{2}-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(12)^{2} )=-28.18x10^{4}J[/tex]

d) The change in kinetic energy where the two vehicles stopped in the collision is:

ΔK=[tex]k_{2}-k_{1}=0-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]

ΔK=[tex]-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(2.15)^{2} )=-10.33x10^4J[/tex]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

[tex]d=v\times t[/tex]

[tex]d=340\ m/s\times 5\ s[/tex]  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

Answer:

9.4 m

Explanation:

We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.

If the ballon is launched at 9.7 m/s at 39 degrees of elevation.

Vx0 = 9.7 * cos(39) = 7.5 m/s

Vy0 = 9.7 * sin(39) = 6.1 m/s

If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

a = -9.81 m/s^2

It will fall when Y(t) = 0

0 = 6.1 * t - 4.9 * t^2

0 = t * (6.1 - 4.9 * t)

t1 = 0 (this is when the balloon was launched)

0 = 6.1 - 4.9 * t2

4.9 * t2 = 6.1

t2 = 6.1 / 4.9 = 1.25 s

The distance from the car will be the horizonta distance it travelled in that time

X(t) = X0 + Vx0 * t

X(1.25) = 7.5 * 1.25 = 9.4 m

Answer:

The kinetic energies are 0.46 eV and 0.78 eV.

Explanation:

Given that,

Work function = 1.5 eV

Wavelength = 632.8 nm

We need to calculate the energy for red helium neon laser

Using formula of energy

[tex]E=\dfrac{hc}{\lambda}[/tex]

Put the value into the formula

[tex]E=\dfrac{4.136\times10^{-15}\times3\times10^{8}}{632.8\times10^{-9}}[/tex]

[tex]E=1.96\ eV[/tex]

We need to calculate the kinetic energy

Using formula of K.E

[tex]K.E=E-\phi[/tex]

[tex]\phi[/tex]=work function

E = energy

Put the value into the formula

[tex]K.E=1.96-1.5[/tex]

[tex]K.E=0.46\ eV[/tex]

We need to calculate the energy for green helium neon laser

Using formula of energy

[tex]E=\dfrac{4.136\times10^{-15}\times3\times10^{8}}{543.5\times10^{-9}}[/tex]

[tex]E=2.28\ eV[/tex]

We need to calculate the kinetic energy

Using formula of K.E

[tex]K.E=2.28-1.5[/tex]

[tex]K.E=0.78\ eV[/tex]

Hence, The kinetic energies are 0.46 eV and 0.78 eV.

The magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.

The vertical displacement of the ink drop as it passes between the deflecting plates is due to the electric force acting on the charged drop. The force of gravity on the drop is negligible compared to the electric force.

Given:

Mass of the ink drop [tex](m) = 1.00 \times 10^{-11}\ kg[/tex]

Vertical displacement[tex](d) = 0.290\ mm[/tex]

Horizontal velocity [tex](v) = 23.0\ m/s[/tex]

Electric field magnitude [tex](E) = 8.20 \times 10^4\ N/C[/tex]

Plate length [tex](D_o) = 1.80\ cm[/tex]

The electric force on a charged particle in an electric field is given by Coulomb's law:

[tex]F_{electric} = q \times E[/tex]

Where:

[tex]F_{electric}[/tex] is the electric force,

q is the charge of the particle,

E is the magnitude of the electric field.

The vertical displacement (d) of the drop is caused by this electric force and is given by the equation:

[tex]d = (1/2) \times a \times t^2[/tex]

Since the drop is initially moving horizontally and is not influenced by gravity, the vertical acceleration (a) is due solely to the electric force

[tex]F_{electric} = m \times a[/tex]

Equate the two expressions for the electric force:

[tex]q \times E = m \times a\\a = (q \times E) / m\\d = (1/2) \times ((q \times E) / m) \times t^2[/tex]

The time it takes to travel through the plates using the horizontal distance [tex](D_o)[/tex] and the horizontal velocity:

[tex]t = D_o/ v[/tex]

[tex]d = (1/2) \times ((q \times E) / m) \times (D_o / v)^2\\q = (2 \times m \times d \times v^2) / (E \times D_o^2)\\q = (2 \times (1.00 \times 10^{-11}) \times (0.290 \times 10^{-3} ) \times (23.0 )^2) / ((8.20 \times 10^4) \times (1.80 \times 10^{-2})^2)\\q = 7.5 \times 10^{-14}[/tex]

Therefore, the magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.

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To find the charge necessary for an ink drop to be deflected a certain distance in an electric field, apply the equation q = Fe/E, where Fe is derived from the kinematic equations for the ink drop's motion in the field.

To calculate the magnitude of charge q that must be given to an ink drop such that it is deflected by a distance d = 0.290 mm by the time it reaches the end of the deflection plate of length D0 = 1.80 cm in a uniform electric field with magnitude E = 8.20×104 N/C, we can use the concepts of electric force and motion.

Firstly, the electric force Fe acting on the ink drop is equal to qE. This force will cause the ink drop to accelerate. Since the drops are moving horizontally at velocity v = 23.0 m/s and they need to be deflected by a certain distance d while travelling through the electric field, we can use the kinematic equations of motion to find the vertical displacement due to acceleration.

The time t it takes for the ink drop to pass through the plates is t = D0/v. From this, the vertical displacement d can be given as d = 0.5 * a * t^2 where a is the acceleration due to the electric force. Knowing d, t, and E, we can find a and subsequently use Fe = mdrop × a to solve for the charge q as q = Fe/E.

Using the given mass m = 1.00×10−11 kg for the ink drop, we can rearrange and solve the appropriate equations to find the value of the charge q necessary to cause the required deflection d.

Answer:

Rate of heat loss=7992 kJ/h

Explanation:

First, we can consider the room as a closed system, so we can use the first law of the thermodynamic:

[tex]Qnet,in-Wnet,out = Esystem[/tex]

The incoming net heat is the incoming heat less the leaving heat, the leaving net work is the leaving work less the incoming work  and the system energy is only the change of the intern energy because we have a stationary system, so:

[tex](Qin-Qout)-(Wout-Win) =U2-U1[/tex]

From the problem we consider that there is no change of the temperature thus there is no change in the intern energy(U2=U1), moreover there is no incoming heat neither exit work, all the work is made to the system as a result the leaving heat is equal to the incoming work:

[tex](0-Qout)-(0-Win) =0[/tex]

[tex]-Qout+Win =0[/tex]

[tex]Qout=Win[/tex]

Substituting the known values we can get the rate of heat loss (exit heat):

[tex]Q_{out}=Q_{loss}=W_{ref}+W_{TV}+W_{res}+W_{fan}[/tex]

[tex]Q_{loss}=250 [W]+120 [W]+1800 [W]+50 [W][/tex]

[tex]Q_{loss}=2220 [W]=2220 [\frac{J}{s} ][/tex]

Converting to Joules/hours:

[tex]Q_{loss}=2220 [\frac{J}{s} ][\frac{3600 s}{1 h} ][/tex]

[tex]Q_{loss}=7992000 [\frac{J}{h} ][/tex]

Finally the rate of heat loss is:

[tex]Q_{loss}=7992 [\frac{kJ}{h} ][/tex]

Final answer:

The rate of heat loss from a room with several electrical appliances running and constant air temperature is 7992 kJ/h, calculated by summing the power of all appliances and converting to kJ/h.

Explanation:

To find the rate of heat loss from the room, we need to consider the energy input from all the electrical appliances and the fact that the air temperature in the room remains constant. The total power consumption of the appliances is the sum of the power of the refrigerator, TV, electric resistance heater, and fan, which is 250 W + 120 W + 1.8 kW + 50 W = 2.22 kW.

Since the temperature is constant, the rate of energy input equals the rate of heat loss. This input is the combined power consumption of the appliances and is entirely converted into heat (assuming 100% efficiency for simplicity).

To express this rate in kilojoules per hour (kJ/h), multiply the power in kilowatts by the number of hours and by 1000 to convert from kW to kJ since 1 kW = 1 kJ/s:

2.22 kW = 2.22 kJ/s

So, the rate of heat loss is:

2.22 kJ/s × 3600 s/h = 7992 kJ/h

Answer:

7.468 kN

Explanation:

Here the force of 7468 Newton is given.

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

The number is 7468.0

Here, the only solution where the number of significant figures is kilo

1 kilonewton = 1000 Newton

[tex]1\ Newton=\frac{1}{1000}\ kilonewton[/tex]

[tex]\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton[/tex]

So 7468 N = 7.468 kN

Answer:

the greatest height at which the ball can reach is 19.62 m.

Explanation:

given,

time taken by the ball to reach the ground = 4 s

time at which ball reach at maximum height = 4/2 = 2 s

velocity of the ball at the top most point = 0 m/s

we know,

v = u + a t

0 = u + (-9.81 ) × 2

u = 19.62 m/s

maximum height achieved

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

[tex]s = 19.62\times 2 + \dfrac{1}{2}(-9.81)\times 2^2[/tex]

s = 19.62 m

hence, the greatest height at which the ball can reach is 19.62 m.

Answer:

Explanation:

Given

Distance to grandmother's house=100 mi

it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph

Let Julie travel x mi with 30 mph and 100-x with 70 mph

[tex]\frac{x}{30}=\frac{100-x}{70}[/tex]

x=30 mi

Therefore

Julie's Average speed on the way to Grandmother's house[tex]=\frac{100}{\frac{50}{30}+\frac{50}{70}}[/tex]

=42 mph

On return trip

[tex]=\frac{100}{2\frac{30}{30}}=50 mph[/tex]

Final answer:

Julie's average speed to Grandmother's house is 42.02 mph. The average speed for the return trip cannot be calculated without the total time of travel or the division of time at each speed. Her average speed for the entire round trip is approximately 26.67 mph, and the average velocity is 0 mph since the starting and ending points are the same.

Explanation:

To calculate Julie's average speed on the way to Grandmother's house, we need to use the formula for average speed, which is total distance traveled divided by the total time taken. Since Julie drives half the distance at 30.0 mph and the other half at 70.0 mph for a total distance of 100 miles, we can calculate the time taken for each half. At 30 mph, for 50 miles, the time taken is ≈ 1.67 hours, and at 70 mph, for 50 miles, the time taken is ≈ 0.71 hours. The total time is ≈ 2.38 hours. Therefore, average speed is 100 miles ÷ 2.38 hours = 42.02 mph.

On the return trip, she drives half the time at 30.0 mph and half the time at 70.0 mph. However, without additional information, we cannot calculate the average speed for the return trip because we need the total time or the portion of time spent at each speed. The calculation is different from the first trip because this time it is dependent on time, not distance. For her entire trip, if Julie returned home 7 hours and 30 minutes after she left, and assuming the same 100-mile distance back, her average speed for the entire trip is the total distance (200 miles) divided by the total time (7.5 hours), which is ≈ 26.67 mph. However, since she returns to the starting point, her displacement is zero, and thus her average velocity for the entire trip is 0 mph, similar to the example in Figure 2.10.

Answer:

the magnitude of the force is 192.29 N

Solution:

As per the question:

Charges present on the corner of the triangle are same, Q = [tex]- 3.1\times 10^{- 9} C[/tex]

Since, its an equilateral triangle, distance between the charges, l = 0.50 m

Now,

The Coulomb force on a charge  due to the other is:

[tex]F_{C} = K\frac{Q^{2}}{l^{2}}[/tex]

where

K = Coulomb constant = [tex]9\times 10^{9} C^{2}/m^{2}[/tex]

[tex]F_{C} = (9\times 10^{9})\frac{(3.1\times 10^{- 9})^{2}}{0.5^{2}}[/tex]

[tex]F_{C} = 111.6 N[/tex]

The the net force on the charges in an equilateral triangle on all the charges due to each other:

[tex]F_{eq} = \sqrt{3}F_{C} = \sqrt{3}\times 111.6 = 193.29 N[/tex]

The magnitude of the force on each charge of -3.1 x 10^-9 C at the corners of an equilateral triangle 0.50 m on a side is 1.088 x 10^-7 N.

For one charge, the force due to the other two charges will be the vector sum of the forces due to each charge separately. Since the triangle is equilateral, the forces will have the same magnitude but different directions. Using Coulomb's law, we find the magnitude of the force between any two charges:

F = k × |q1 × q2| / r2

Where:

Substituting the given values:

F = (8.988 x 109 Nm2/C2) × (3.1 x 10-9 C)2 / (0.50 m)2

This results in:

F = 1.088 x 10-7 N

As all sides are equal, and charges are equal, the net force will be the same in magnitude for each charge. The direction of the resulting force on each charge will be along the perpendicular bisector of the side opposite to each charge due to the symmetry of the configuration.

Answer:

13.65 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 9.5 m

a = Acceleration = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 9.5-0^2}\\\Rightarrow v=13.65\ m/s[/tex]

The flower pot is moving at a speed of 13.65 m/s when it crashes the sidewalk.

Answer:

Distance, d = 101.388 meters

Explanation:

It is given that,

The average velocity of plane, v = 3650 km/h = 1013.88 m/s

The time for which eye blinks, [tex]t=100\ ms=0.1\ s[/tex]

Let d is the distance covered by the jet. It can be calculated as :

[tex]d=v\times t[/tex]

[tex]d=1013.88\ m/s\times 0.1\ s[/tex]        

d = 101.388 meters

So, the distance covered by a fighter jet during a pilot's blink is 101.388 meters. Hence, this is the required solution.                                                                              

Final answer:

During a blink that lasts 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.

Explanation:

To calculate the distance a fighter jet travels during a pilot's blink, we can use the formula distance = velocity × time. The pilot's blink lasts 100 milliseconds (ms) which is 0.1 seconds because 1000 ms equals 1 second. The jet's velocity is given as 3650 kilometers per hour (km/h).

First, we'll need to convert the jet's velocity to meters per second (m/s) since the time of the blink is given in seconds. 1 km equals 1000 meters, and there are 3600 seconds in an hour, so:

Velocity in m/s = (Velocity in km/h) × (1000 m/km) / (3600 s/h) = (3650) × (1000) / (3600) = 1013.89 m/s

Now, we will multiply the velocity in m/s by the time in seconds to get the distance:

Distance = Velocity × Time = (1013.89 m/s) × (0.1 s) = 101.39 meters

So, during the blink of an eye, which is 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.

Answer:

260 g

Explanation:

Given:

Let n be the number of such silicon dioxide that would give the desired mass of sand.

According to the question, the surface area of all the sand particles will be equal to the surface area of the cube.

[tex]\therefore n\times \textrm{Area of a sand particle}=\textrm{Surface area of a cube }\\\Rightarrow n\times 4\pi r^2= 6a^2\\\Rightarrow n = \dfrac{6a^2}{4\pi r^2}[/tex]

Let the total mass of sand required be M.

[tex]\textrm{Total mass of sand} = \textrm{Mass of all the required sand particle}\\\Rightarrow M = n\times \textrm{mass of one sand particle}\\\Rightarrow M = n\times \textrm{Density of sand particle}\times \textrm{Volume of a sand particle}\\\Rightarrow M = n\times\rho \times \dfrac{4}{3}\pi r^3\\[/tex]

[tex]\Rightarrow M = \dfrac{6a^2}{4\pi r^2}\times\rho \times \dfrac{4}{3}\pi r^3\\\Rightarrow M = 2\times\rho \times a^2r\\\Rightarrow M = 2\times2600 \times (1)^2\times 5\times 10^{-5}\\\Rightarrow M =0.260\ kg\\\Rightarrow M =260\ g\\[/tex]

Hence, the total mass of the sand required will be equal to the 260 g.

Answer:

mass = 61.16 kg

Explanation:

given data

distance = 50 m

energy = 500 W = 500 × 60sec = 30000 J/min

to find out

mass

solution

we will apply here energy equation  

that is

energy = m×g×h    .............1

put here all value m mass here and g = 9.81 and h is distance

energy = m×g×h

30000 = m×9.81×50

mass = 61.16 kg

Answer:

61 kg

Explanation:

Given data

distance (d): 50 m

power (P): 500 W = 500 J/s

mass (m): ?

time (t): 1 min = 60 s

The engine must do some work (w) to lift the coal. The work done in 1 minute is:

P = w / t

w = P . t = 500 J/s × 60 s = 3.0 × 10⁴ J

The work is equal to the force exerted (F) times the distance (d).

w = F × d

F = w / d = 3.0 × 10⁴ J / 50 m = 6.0 × 10² N

The force exerted is equal to the mass lifted (m) times the acceleration. Here, the acceleration to surpass is that of gravity (g = 9.8 m/s²).

F = m × g

m = F / g = 6.0 × 10² N / 9.8 m/s² = 61 kg

The sound would reach the dolphin in the water about 2.12 seconds earlier than the person in the boat, using given speeds of sound in air and water, and the distance provided.

The time it takes for sound to travel is calculated using the formula Time = Distance/Speed. Given, distance is 1000m in both air and sea water. We need to find the speed of sound in the air to calculate how long it will take the sound to reach the person in the boat.

The speed of sound in the air at 25°C is approximately 346.13 m/s. Using the formula: Time = Distance/Speed, the time it takes for the sound to reach the person in the boat is about 2.889 seconds (1000m/346.13 m/s).

The speed of sound in seawater is given as 1,533 m/s. Using the same formula, the time it takes for the sound to reach the dolphin in the water is about 0.652 seconds (1000m/1,533 m/s).

Therefore, the difference in time it takes for the sound to travel to the person and the dolphin is approximately 2.237 seconds (2.889 seconds - 0.652 seconds). Hence, the correct answer would be (B) 2.12. Although, it's close to the calculated difference, none of the available options are an exact match.

https://brainly.com/question/35989321

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Answer: 0.506 m

Explanation: To solve this we use the relationship for the harmonic in the string which are given by the following expression:

λ=2*L/3

λ=2*0.76 m/3= 0.506 m

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

Final answer:

The correct statement is that c) at the highest point of its trajectory, a rock thrown vertically has a velocity of zero and an acceleration directed downward at -9.80 m/s². The acceleration due to gravity is constant throughout the motion, whether the rock is rising or falling.

Explanation:

Among the given statements regarding a rock thrown vertically upward and neglecting air resistance, the true statement is: c) At the highest point the velocity is zero, the acceleration is directed downward. This is because at the maximum height, the rock stops for an instant (velocity = 0 m/s), but the acceleration due to gravity (acceleration due to gravity) is still acting on it, which is approximately -9.80 m/s² towards the Earth. This acceleration remains constant whether the rock is moving up or down.

It's important to note that as the rock rises, its acceleration vector always points downward, opposite to the direction of its velocity. Once the rock reaches its maximum height and begins its descent, the velocity becomes negative, which indicates the rock is falling towards the ground, yet still the acceleration is -9.80 m/s² downwards. The velocity and acceleration are in the same direction during the fall, both pointing toward the Earth.

Answer:

The cannon ball hits the ground with a speed of 76.35 m/s.

Explanation:

We shall use the conservation of energy principle to solve the problem

The initial energy of the cannon ball is the sum of the potential and the kinetic energies.

We know that potential energy = [tex]mgh[/tex]

Applying the given values we obtain the initial potential energy as

[tex]P.E=m\times g\times 18.0[/tex]

Similarly the initial kinetic energy of the ball equals

[tex]K.E_{initial}=\frac{1}{2}mv_{o}^{2}[/tex]

Applying values we get

[tex]K.E_{initial}=\frac{1}{2}m(74.0)^{2}[/tex]

Thus the initial energy is thus [tex]E_{initial}=m\times g\times 18.0+\frac{1}{2}\times m\times (74.0)^{2}[/tex]

Now upon hitting the ground the only energy that the cannon ball will posses is kinetic energy since the potential energy of any object upon touching the surface of earth equals zero

Thus we have

[tex]Energy_{final}=\frac{1}{2}mv_{f}^{2}[/tex]

Equating initial and final energies we get

[tex]m\times g\times 18+\frac{1}{2}\times m\times (74.0)^{2}=\frac{1}{2}\times m\times v_{f}^{2}\\\\v_{f}^{2}=36g+(74.0)^{2}\\\\\therefore v_f=\sqrt{36\times 9.81+(74.0)^{2}}\\\\v_f=76.35m/s[/tex]

Answers:

a) [tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]

b) [tex]\rho_{liq}= 1.48 g/cm^{3}[/tex]

c) When we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.

Explanation:

a) According to Archimedes’ Principle:

A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.

In this case we have a wooden cylinder floating (partially immersed) in water. This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium.  This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.

Hence:

[tex]W_(cylinder)=B[/tex] (1)

Where:

[tex]W_(cylinder)=m.g[/tex] is the weight of the wooden cylinder, where [tex]m[/tex] is its mass and [tex]g[/tex] gravity.

[tex]B[/tex] is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.

We can rewrite (1) as follows:

[tex]m_{cylinder}g=m_{water}g[/tex] (2)

On the other hand, we know density [tex]\rho[/tex] establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:

[tex]\rho=\frac{m}{V}[/tex] (3)

isolating the mass:

[tex]m=\rho V[/tex]    (4)

Now we can express (2) in terms of the density and the volume of cylinder and water:

[tex]\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g[/tex] (5)

In this case [tex]V_{water}[/tex] is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).

At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:

[tex]V_{cylinder}=\pi r^{2} h[/tex] (6)

Where [tex]r[/tex] is the radius and [tex]h=30 cm[/tex] the total height of the cylinder.

[tex]V_{water}=\pi r^{2} (h-h_{top})[/tex] (7)

Where [tex]h_{top}=13.5 cm[/tex] is the height of the top of the cylinder above the surface of water and [tex](h-h_{top})[/tex] is the height of the sumerged part of the cylinder.

Substituting (6) and (7) in (5):

[tex]\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g[/tex] (8)

Clearing [tex]\rho_{cylinder}[/tex]:

[tex]\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}[/tex] (9)

Simplifying;

[tex]\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}[/tex] (10)

Knowing [tex]\rho_{water}=1g/cm^{3}[/tex]:

[tex]\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})[/tex] (11)

[tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex] (12) This is the density of the wooden cylinder

b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ([tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]), but the density of the liquid [tex]\rho_{liq}[/tex] is unknown.

Applying again the Archimedes principle:

[tex]\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g[/tex] (13)

Isolating [tex]\rho_{liq}[/tex]:

[tex]\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}[/tex] (14)

Where:

[tex]V_{cylinder}=\pi r^{2} h[/tex]

[tex]V_{liq}=\pi r^{2} (h-h_{top})[/tex]

Then:

[tex]\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}[/tex] (15)

[tex]\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}[/tex] (16)

[tex]\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}[/tex] (17)

[tex]\rho_{liq}= 1.48 g/cm^{3}[/tex] (18) This is the density of the liquid

c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.

This is because in this case, when we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.

Explanation:

Charge, [tex]q=-30\ \mu C=-30\times 10^{-6}\ C[/tex]

Electric force, [tex]F=27\ mN=27\times 10^{-3}\ N[/tex]

We need to find the magnitude and direction of electric field at that location. The relation between the electric field and electric force is given by :

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{27\times 10^{-3}\ N}{-30\times 10^{-6}\ C}[/tex]

[tex]E=-900\ N/C[/tex]

For a negative charge, the direction of electric field is inward. The direction of electric force and electric field is same. So, the direction of electric field in this case is in upward direction. Hence, this is the required solution.

The magnitude of the electric field is 900 N/C, and the direction is downward. Therefore, the electric field at that location is  -900 N/C  (negative value indicating downward).

To determine the magnitude and direction of the electric field at the location where the point charge experiences an electrostatic force, we can use the relationship between the force  [tex]\mathbf{F}[/tex]  experienced by a charge  q  in an electric field [tex]\mathbf{E}[/tex] :

[tex]\mathbf{F} = q \mathbf{E}[/tex]

Given:

- The charge  q  is -30 µC (microcoulombs).

- The electrostatic force  [tex]\mathbf{F}[/tex]  is 27 mN (millinewtons) upward.

First, we convert the given quantities to standard SI units:

- Charge  [tex]q = -30 \, \mu \text{C} = -30 \times 10^{-6} \, \text{C}[/tex]

- Force  [tex]\mathbf{F} = 27 \, \text{mN} = 27 \times 10^{-3} \, \text{N}[/tex]

Next, we use the equation  [tex]\mathbf{F} = q \mathbf{E}[/tex] to solve for the electric field [tex]\mathbf{E}[/tex] :

[tex]\mathbf{E} = \frac{\mathbf{F}}{q}[/tex]

Substituting the given values:

[tex]\mathbf{E} = \frac{27 \times 10^{-3} \, \text{N}}{-30 \times 10^{-6} \, \text{C}} \\\\\mathbf{E} = \frac{27 \times 10^{-3}}{-30 \times 10^{-6}} \, \text{N/C} \\\\\mathbf{E} = \frac{27}{-30} \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -0.9 \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -900 \, \text{N/C}[/tex]

The negative sign indicates the direction of the electric field. Since the force on a negative charge is upward, the electric field must be directed downward (opposite to the direction of the force on a negative charge).

Final answer:

To find the angle and tension in the wire, we can use the concept of electrostatic force and Coulomb's Law.

Explanation:

To find the angle and tension in the wire, we can use the concept of electrostatic force. The electrostatic force between the two charges can be calculated using Coulomb's Law:

FE = k * (q1 * q2) / r^2

where FE is the electrostatic force, k is the electrostatic constant (8.99 x 10^9 N·m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

By equating this force to the tension in the wire and solving for the angle, we can find the angle:

tan(angle) = FE / T

where T is the tension in the wire.

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

[tex]1\ m=\frac{1}{1609.344}\ miles[/tex]

1 hour = 60×60 seconds

[tex]1\ s=\frac{1}{3600}\ hours[/tex]

[tex]28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h[/tex]

Top speed of the cheetah is 64.2 mi/h

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2[/tex]

Acceleration of the cheetah is 8.68 m/s²

2)

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s[/tex]

It takes a cheetah 3.31 seconds to reach its top speed.

3)

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m[/tex]

It travels 47.5 m in that time

4) When s = 120 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s[/tex]

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

The cheetah's top speed of 28.7 m/s is approximately 64.2 mi/h. It takes a cheetah 3.31 seconds to reach its top speed, traveling a distance of 47.4 meters during this acceleration. To reach a stationary rabbit 120 meters away, it would take the cheetah a total of 5.84 seconds.

The question involves converting speeds from meters per second to miles per hour, finding the time taken to achieve a certain speed, calculating the distance traveled in that time, and determining the time required to reach a target.

Answer:

a) [tex]R_s = 0.092[/tex]

b) [tex]R_p = 0.085[/tex]

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

[tex]R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2[/tex]

using snell's law to calculate θ t

[tex]sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}[/tex]

[tex]cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}[/tex]

a) [tex]R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2[/tex]

[tex]R_s = 0.092[/tex]

b) [tex]R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2[/tex]

[tex]R_p = 0.085[/tex]

Answer:

0.0567 N

Explanation:

q1 = 3 micro coulomb

q2 = - 2 micro coulomb

OB = 50 cm

AB = 60 cm

By using Pythagoras theorem in triangle OAB

[tex]OA^{2}=OB^{2}+AB^{2}[/tex]

[tex]OA^{2}=50^{2}+60^{2}=6100[/tex]

OA = 78.1 cm

By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is

[tex]F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}[/tex]

F = 0.0885 N

The horizontal component of force is

= F CosФ = [tex]F\times \frac{OB}{OA}[/tex]

= 0.0885 x 50 / 78.1 = 0.0567 N