A rifle with a mass of 0.9 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 750 m/s after the rifle is fired a. What is the momentum of the bullet after the rifle is fired? b. If external forces acting on the rifle can be ignored, what is the recoil velocity of the rifle?

Answer:

490hp

Explanation:

Power is energy per unit of time, an in this case the energy needed is the gravitational potential energy, so we have:

[tex]P=E/t=mgh/t=(3730kg)(9.8m/s)(600m)/60s=365540W[/tex]

Since all units are in S.I. we get our result in Watts (Joules/s). To convert to horse power (imperial), we need to know that:

[tex]745.7 W = 1 hp[/tex]

Which obviously means:

[tex]\frac{1 hp}{745.7 W} = 1[/tex]

So we can write:

[tex]P=365540W=365540W(\frac{1 hp_m}{745.7 W})=490hp[/tex]

There is also metric horse power ([tex]735.5 W = 1 hp[/tex]), and using this value we get [tex]P=497hp[/tex], although both results are close.

Final answer:

To calculate the SUV's power output, we find the work done against gravity and divide it by the time taken to get power in watts. This is then converted to horsepowers leading to the answer of 490 hp. The correct answer is b) 490 hp.

Explanation:

To find the power output in horsepowers of the 3730-kg SUV climbing a 600-meter hill in 60 seconds, first we calculate the work done against gravity, which is equal to the force due to gravity (weight of the SUV) multiplied by the height of the hill:

Work = Weight × Height = (mass × gravity) × height

Then, we convert the weight to newtons by multiplying the mass (kg) by the acceleration due to gravity (9.81 m/s2, approximately), and multiply that result by the height (in meters).

Power in watts is calculated by dividing the work done by the time taken in seconds:

Power (W) = Work done (Joules) / Time (seconds)

To convert watts to horsepowers, we use the conversion rate where 1 horsepower is equivalent to 746 watts:

Power (hp) = Power (W) / 746

After plugging in the values, the SUV's power output in horsepowers would be:

Power (W) = (3730 kg × 9.81 m/s2) × 600 m / 60 s = 365,130 W

Power (hp) = 365,130 W / 746 = 489.43 hp, which we round to 490 hp.

Thus, the correct answer from the given options is 490 hp.

Answer:

The pressure of the tire will be 4.19 atm

Explanation:

The ideal gas equation is:

p0 * V0 / T0 = p1 * V1 / T1

Since V0 = V1 because the volume of the tire doesn't change

p0 / T0 = p1 / T1

p1 = p0 * T1 / T0

We need absolute temperatures for this equation

16 C = 289 K

45 C = 318 K

Now we replace the values:

p1 = 3.71 * 318 / 289 = 4.19 atm

Final answer:

To calculate the new gauge pressure of a heated bicycle tire with constant volume, we use Charles's law. With initial conditions of 3.71 atm and 16°C, and final temperature 45°C, the final gauge pressure is approximately 4.08 atm.

Explanation:

The student is asking to calculate the new gauge pressure of a bicycle tire after its temperature increases, assuming the volume and amount of gas remain constant.

This scenario is governed by the Gas Law, specifically Charles's law, which states that the pressure of a gas is directly proportional to its temperature when the volume is held constant.

Therefore, one can apply the formula P2 = P1 * (T2 / T1), where P1 and T1 represent the original pressure and temperature and P2 and T2 represent the final pressure and temperature, respectively, to calculate the new pressure at the higher temperature.

Considering the initial gauge pressure (P1) is 3.71 atm and the initial temperature (T1) is 16°C or 289 K (since temperatures must be in Kelvin for these calculations), and the final temperature (T2) is 45°C or 318 K, we can solve for P2.

First, add 273.15 to the temperature to convert it from Celsius to Kelvin.

T1 = 16 + 273.15 = 289.15 K

T2 = 45 + 273.15 = 318.15 K

Next, apply Charles's law to find P2:

P2 = P1 * (T2 / T1)

P2 = 3.71 atm * (318.15 K / 289.15 K)

P2 = 3.71 atm * 1.1004

P2 ≈ 4.08 atm

The final gauge pressure in the tire at the higher temperature of 45°C is approximately 4.08 atm.

Answer:

Number of gallons =2 gallon

Explanation:

given data:

rate of gasoline ineurope = 5 euro per liter

total money to buy gasoline =  40 euro

total gasoline an american can buy in europe [tex]= \frac{40}{5}[/tex]

= 8 litres of gasoline

As given in the question 1 ltr is 1 quarts therefore  

Total no. of quarts is 8 quarts

As from question 4 quarts is equal to one gallon, hence

Number of gallons[tex]= \frac{8}{4} = 2 gallon[/tex]

Answer:

39.56 x 10⁻² V

Explanation:

Peak emf in a rotating coil

= n BAω

Where n is no of turns of coil , B is magnetic field , A is area of coil and ω is angular velocity of rotation of coil

Here n = 1

B = .21 T

A = 5 X 12 X 10⁻⁴ = 60 X 10⁻⁴

ω = 2π X 50 = 100π

Emf = 1 x .21 x 60 x 10⁻⁴ x 100 x 3.14

=  39.56 x 10⁻² V

Answer:

[tex]3.75*10^{-13}[/tex]  electrons

Explanation:

The total charge Q is the sum of the charge of the N electrons contained in the sphere:

[tex]Q=N*q_{e}[/tex]

[tex]q_{e}=-1.6*10^{-19}C[/tex]    charge of a electron

We solve to find N:

[tex]N=\frac{Q}{q_{e}}=\frac{-6 x 10^{-6}}{-1.6 x 10^{-19}}=3.75*10^{-13}[/tex]

Answer:

The correct option is 'c': Decreases.

Explanation:

As we know that when we throw a ball upwards we imoart a kinetic energy to the bsll. As the ball moves upwards the kinetic energy of the ball starts to decrease as the ball slows down as it moves upwards as work has to be done for movement against gravity. This work done against gravity is stored as potential energy of the ball.

Mathematically as we throw the ball upward's with velocity 'u' it's initial kinetic energy equals [tex]E_{initial}=\frac{1}{2}mu^{2}[/tex]

As the ball attain's a height 'h' it's total energy is sum of the potential and the remaining kinetic energy as follows

[tex]E(h)=mgh+\frac{1}{2}mv^{2}[/tex]

Equating both the energies we get

[tex]\frac{1}{2}mu^{2}=mgh+\frac{1}{2}mv^{2}\\\\\\therefore v=\sqrt{u^{2}-2gh}[/tex]

Hence we can see as the height increases the velocity decreases.

Answer: Hi!

The acceleration of our particle is then a = 4*[tex]10^{3[/tex] meters per second square. Also we know that we start at X₀ = O m (i am not sure if you are using a a 0 or a O, but you will see that in this problem it doesn't matter, se i will use a constant O, that can be  ani real number as the initial position) and V₀ = 0 m/s.

Ok, we have the acceleration, so if we want the velocity, we must integrate the acceleration over time.

V(t) = ∫ 4*[tex]10^{3[/tex]dt = 4*[tex]10^{3[/tex]*t + V₀

where we added the constant of integration, who is the inicial velocity, that we already know that is zero.

and for the position, we must integrate again.

X(t) =  ∫ 4*[tex]10^{3[/tex]*tdt =  4*[tex]10^{3[/tex]*[tex]t^{2}[/tex]/2 + X₀

ok, in the part a) we want to know the displacement at 2 seconds.

that is X(2s) - X(0s) = (2000*4 + O) - O meters = 8000 meters.

in the b part we want to know the velocity at 1 second; that is:

V(1s) = 4000*1 m/s = 4000 meters per second.

Answer:

The average speed is less than 70 km/h.

(B) is correct option.

Explanation:

Given that,

distance = 6.0 km

Speed = 50 km/h

Speed = 90 km/h

We need to calculate the time in 6.0 km distance

Using formula of time

[tex]t = \dfrac{d}{v}[/tex]

Put the value in to the formula

[tex]t=\dfrac{6.0}{50}[/tex]

[tex]t=0.12\ hr[/tex]

We need to calculate the time in another distance

Using formula of time

[tex]t = \dfrac{d}{v}[/tex]

Put the value in to the formula

[tex]t=\dfrac{6.0}{90}[/tex]

[tex]t=0.067\ hr[/tex]

We need to calculate the average speed

Using formula of average speed

[tex]v=\dfrac{D}{T}[/tex]

Where, D = total distance

T = total time

Put the value into the formula

[tex]v=\dfrac{12}{0.12+0.067}[/tex]

[tex]v=64.17\ km/hr[/tex]

Hence, The average speed is less than 70 km/h.

The average speed over the 12 km drive is less than 70 km/h since the total time for the trip is 0.187 hours, leading to an average speed of 64.2 km/h.

To calculate the average speed for the entire trip, you must divide the total distance by the total time taken. In the given scenario, you drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. First, calculate the time taken for each portion of the trip.

Now, add the times together and divide the total distance by the total time to find the average speed.

Therefore, the average speed over the 12 km drive is less than 70 km/h, which corresponds to option (B).

Answer:

acceleration = 0.8181 m/s²

Explanation:

given data

mass = 1.1 kg

apart d = 1 m

charge q = 10 μC

to find out

What is the initial acceleration

solution

we know that acceleration is

acceleration = [tex]\frac{force}{mass}[/tex]   .................1

here force = [tex]k \frac{q1q2}{r^2}[/tex]

here q1 q2 is charge and r is distance and Coulomb constant k = 9 × [tex]10^{9}[/tex] Nm²/C²

force = [tex]9*10^{9} \frac{(10*10^{-6})^2}{1^2}[/tex]

force = 0.9 N

so  from equation 1

acceleration = [tex]\frac{0.9}{1.1}[/tex]

acceleration = 0.8181 m/s²

Answer:

21 m/s

Explanation:

For the first ball, in the x direction:

x = x₀ + v₀ t + ½ at²

x = 0 + (8.2 cos 35) t + ½ (0) t²

x = 6.72t

In the y direction:

y = y₀ + v₀ t + ½ at²

y = 8 + (8.2 sin 35) t + ½ (-9.8) t²

y = 8 + 4.70t − 4.9t²

When y = 4:

4 = 8 + 4.70t − 4.9t²

4.9t² − 4.70t − 4 = 0

Solve for t with quadratic formula:

t = [ 4.70 ± √((-4.70)² − 4(4.9)(-4)) ] / 9.8

t = (4.70 ± 10.0) / 9.8

t = 1.50

Therefore:

x = 6.72t

x = 10.1

Now, for the second ball in the x direction:

x = x₀ + v₀ t + ½ at²

x = 0 + (v₀ cos (-θ)) (t − 1) + ½ (0) (t − 1)²

x = v₀ cos θ (t − 1)

And in the y direction:

y = y₀ + v₀ t + ½ at²

y = 8 + (v₀ sin (-θ)) (t − 1) + ½ (-9.8) (t − 1)²

y = 8 − v₀ sin θ (t − 1) − 4.9(t − 1)²

When t = 1.50, x = 10.1 and y = 4:

10.1 = v₀ cos θ (1.50 − 1)

v₀ cos θ = 20.1

4 = 8 − v₀ sin θ (1.50 − 1) − 4.9(1.50 − 1)²

4 = 6.76 − 0.50 v₀ sin θ

v₀ sin θ = 5.49

Using Pythagorean theorem:

v₀² = (v₀ cos θ)² + (v₀ sin θ)²

v₀² = (20.1)² + (5.49)²

v₀ = 20.8

Rounded to two significant figures, the required initial speed is 21 m/s.

Answer:

The given statement is false.

Explanation:

For any negative vector

[tex]\overrightarrow{r}=-x\widehat{i}-y\widehat{j}[/tex]

The magnitude of the vector is given by

[tex]|r|=\sqrt{(-x)^{2}+(-y)^{2}}\\\\|r|=\sqrt{x^2+y^2}[/tex]

As we know that square root of any quantity cannot be negative thus we conclude that the right hand term in the above expression cannot be negative hence we conclude that magnitude of any vector cannot be negative.

Answer:

[tex]f_{ecco}  = 360 Hz[/tex]

Explanation:

The change of frecuency of sound due to the movement of the source is colled Doppler Effect.

As James (the source) is running toward the wall, the frecuency reaching the wall (so the eco sound) will be higher than the source. In this case the frecuency at the wall will be:

[tex]f_{2}  = f_{1}  (\frac{v}{v-v_{s} } )[/tex]

where [tex]v_{s}[/tex] is the speed of source, 2 m/s

and [tex]v[/tex] is the speed of sound, given that we have wind movind the air in the opposite direction respect to the wall, the speed of sound would be:

[tex]v = 350 \frac{m}{s} - 3 \frac{m}{s} = 347 \frac{m}{s}[/tex]

Replacing the values: [tex]f_{2}  = 357 Hz[/tex]

Now the wall becames the new source, and James (the observer is aproaching the source), for an observer aproaching the source the new frecuency will be:

[tex]f_{3}  = f_{2}  (1 + \frac{v_{s}}{v} )[/tex]

Now the waves are traveling in the direction of wind, so the velocity of sound will be:

[tex]v = 350 \frac{m}{s} + 3 \frac{m}{s} = 353 \frac{m}{s}[/tex]

Replacing:

[tex]f_{3}  = 360 Hz[/tex]

Answer:

Change in momentum is [tex]2.667\times 10^{4} kg.m/s[/tex]

Solution:

The momentum of any body is the product of its mass and the velocity associated with the body and is generally given by:

[tex]\vec{p} = m\vec{v}[/tex]

Now, as per the question:

Mass of the car, M = 1500 kg

The velocity in the east direction, [tex]v\hat{i} = 40\hat{i} km/h[/tex]

The velocity in the north direction, [tex]v\hat{j} = 50\hat{j} km/h[/tex]

Now, the momentum of the car in the east direction:

[tex]p\hat{i} = mv\hat{i} = 1500\times 40\hat{i} = 60000\hat{i} kg.km/h[/tex]

Now, the momentum of the car in the north direction:

[tex]p\hat{j} = mv\hat{j} = 1500\times 50\hat{j} = 75000\hat{j} kg.km/h[/tex]

Change in momentum is given by:

[tex]\Delta p = p\hat{i} - p\hat{j} = 60000\hat{i} - 75000\hat{j}[/tex]

Now,

[tex]|\Delta p| = |60000\hat{i} - 75000\hat{j}|[/tex]

[tex]|\Delta p| = \sqrt{60000^{2} + 75000^{2}}[/tex]

[tex]|\Delta p| = 96046 kg.km/hr = \frac{96046}{3.6} = 2.667\times 10^{4} kg.m/s[/tex]

(Since, [tex]1kg.km/h = \frac{1}{3.6} kg.m/s[/tex])

Answer:

Speed of ball equals 6.024 m/s.

Explanation:

Let the student throw the ball with a velocity of 'v' m/s horizontally.

Now the time in which the ball travels 10.0 meter horizontally shall be equal to the time in which it travels (15.0-1.50) meters vertically

Hence the time taken to cover a vertical distance of 13.50 meters is obatined using 2 equation of kinematics as

[tex]s=\frac{1}{2}gt^{2}\\\\t=\sqrt{\frac{2s}{g}}\\\\t=\sqrt{\frac{2\times 13.5}{9.81}}\\\\\therefore t=1.66 seconds[/tex]

Since there is no acceleration in horizantal direction we infer that in time of 1.66 seconds the ball travels a distance of 10 meters

Hence the spped of throw is obatines as

[tex]Speed=\frac{Distance}{Time}\\\\v=\frac{10}{1.66}=6.024m/s[/tex]

Final answer:

The initial speed of the softball can be calculated using the height it drops and the horizontal distance it travels. By determining the time of fall for the vertical drop and knowing the horizontal distance, the initial horizontal velocity can be found, approximately 6.02 m/s.

Explanation:

To solve for the initial speed of the softball, we need to analyze the motion in two dimensions separately: horizontal and vertical.

Horizontal Motion

The horizontal motion can be considered with constant velocity since there's no acceleration in that direction (ignoring air resistance).

Vertical Motion

The vertical motion can be described by the kinematic equations for uniformly accelerated motion (free fall). Given the height difference from the window to the catchpoint (13.5 m) and the acceleration due to gravity (9.81 m/s²), we can calculate the time it takes for the ball to fall this height. The equation we use is:

h = vit + (1/2)at²

Substitute h = 13.5 m, a = 9.81 m/s², and vi = 0 m/s (since the ball is thrown horizontally, the initial vertical speed is 0) to solve for t.

We find that the time t is approximately 1.66 seconds. Using this time and the horizontal distance of 10.0 m, we can now calculate the initial horizontal speed.

vh = d/t

Substituting d = 10.0 m and t = 1.66 s, the initial horizontal speed vh is approximately 6.02 m/s.

Answer:

[tex]a=\frac{v_{f}-v_{o}  }{t_{f}-t_{o}  } =\frac{(7-22)\frac{m}{s} }{5s-0s} =-3\frac{m}{s^{2} }[/tex]

Explanation:

We know that the initial velocity is 22 m/s

The final velocity after applying the brakes is 7 m/s, and the time that takes it to break is 5 seconds

We know that the concept of acceleration is the change of velocity in time

[tex]a=\frac{v_{f}-v_{o}  }{t_{f}-t_{o}  } =\frac{(7-22)\frac{m}{s} }{5s-0s} =-3\frac{m}{s^{2} }[/tex]

The answer is negative since the car is slowing down  

Answer:

How: It will have a very different chemical composition.

Why: Because the interstellar medium at the beginning of universe were compounded only of Hydrogen and Helium.

Explanation:

Once the Big Bang passed and the universe began its expansion, the temperature started to decrease, what allowed the combination of hydrogen and helium. Therefore, the chemical composition of the molecular cloud in which the first stars were formed only held those elements.

If the Sun and the solar system were formed from a molecular cloud with that composition, it will both have an absence of metallic elements (elements heavier than helium), since all of them are formed through nuclear reactions in the core of the stars, and each time a star dies, the interstellar medium for the next generation is enriched.

Key elements for life as Carbon, Nitrogen, Oxygen, Phosphorus and Sulfur will be missing for that hypothetical scenario.

For example, the lack of O3 (ozone) in the atmosphere of earth will allow that ultraviolet light hits the earth surface.

If our solar system formed with the first stars, the planets would mostly comprise hydrogen and helium, as heavier elements had not yet formed. The conditions would be hotter and harsher, making life as we know it unlikely to evolve. The Sun, too, would've been exhausted already due to its short lifespan.

If our Sun and the solar system had formed around the same time as the first stars in the universe, our solar system would look very different. The universe at the time of the formation of the first stars was composed of only hydrogen and helium, with trace amounts of lithium. Thus, the planets in our system, formed from the leftover dust and gasses surrounding our Sun, would mostly be made up of hydrogen and helium.

Believed to have occurred just about 200 million years after the big bang, the first stars were much larger and hotter than the current generation of stars, including our Sun. Hence, the planets in our system could have been significantly hotter and larger. It's also worth noting that these first-generation stars had relatively short lifetimes and ended their lives in explosive supernovae, catalyzing the formation of heavier elements. But this means if our Sun was a first-generation star, it would have exhausted its nuclear fuel billions of years ago.

Due to these circumstances, life as we know it might not have been possible in our solar system. Life on Earth relies on heavier elements like carbon, nitrogen, and oxygen, which were only formed in the universe after generations of stellar evolution and supernova explosions. If our solar system was created at the same time as the first stars, these heavier elements wouldn't have existed yet. Therefore, life, at least life as we understand it, wouldn't likely evolve in our solar system.

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Answer: the fraction is [tex]0.88*10^{-6}[/tex]

Explanation:

Hi!

The land area per capita in USA is 28,310 square km. Then the solar power per capita is (rounding some numbers):

[tex]P_s = 300 \frac{W}{m^2} (28,310) km^2 =  3*10^2*2.83*10^4*10^6 W = 8.49*10^{12} W = 8.49*10^9 kW[/tex]

If we take 20% of this power, the fraction k to have 1.5 kW is:

[tex]1.5 kW = k(0.2*8.49 * 10^9 kW)[/tex]

[tex]k = \frac{1.5}{1.7}*10^{-6} = 0.88 *10^{-9} [/tex]

Final answer:

To answer the question, we need to calculate the total electrical energy consumption in the U.S., understand the solar energy per square meter, consider the 20% efficiency of solar cells, and determine the needed land area for solar panels to meet the country's electrical energy needs.

Explanation:

The question involves calculating what fraction of the United States' land area would need to be covered with 20% efficient solar cells to provide all of its electricity, given that the average American uses electricity at the rate of about 1.5 kW and solar energy reaches the Earth's surface at an average rate of about 300 watts per square meter. To find this fraction, we first need to consider the total electrical energy consumption of the United States and then assess how much of this demand can be met by the solar energy available on a given area of land, taking into consideration the efficiency of the solar cells.

The key steps include calculating the total power used by Americans in watts, understanding the energy provided by the sun per square meter, adjusting for the efficiency of solar conversion, and finally, determining the land area required. Specifically, a 20% conversion efficiency means that only a fifth of the solar energy striking the solar cells is converted into usable electrical energy. With these considerations, the solution entails a mix of energy requirement calculations and solar energy potential assessments to identify the required land area for solar panels.

Answer:

[tex]t=6.4534 s[/tex]

Explanation:

This is an exercise where you need to use the concepts of free fall objects

Our knowable variables are initial high, initial velocity and the acceleration due to gravity:

[tex]y_{0}=75m[/tex]

[tex]v_{oy} =20m/s[/tex]

[tex]g=9.8 m/s^{2}[/tex]

At the end of the motion, the rock hits the ground making the final high y=0m

[tex]y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}[/tex]

If we evaluate the equation:

[tex]0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex]

This is a classic form of Quadratic Formula, we can solve it using:

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

[tex]a=-4.9\\b=20\\c=75[/tex]

[tex]t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s[/tex]

[tex]t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s[/tex]

Since the time can not be negative, the reasonable answer is

[tex]t=6.4534s[/tex]

The rock will hit the ground approximately 2.37 seconds after it is thrown.

Given:

- Initial height, [tex]\( y_0 = 75 \)[/tex] m

- Initial velocity, [tex]\( v_0 = 20 \)[/tex] m/s (since the rock is thrown upwards, this velocity will be negative in the equation as it is in the opposite direction to the positive y-axis)

- Acceleration due to gravity, [tex]\( g = 9.8 \) m/s\( ^2 \)[/tex]

The motion equation becomes:

[tex]\[ 0 = 75 - 20t - \frac{1}{2}(9.8)t^2 \][/tex]

Multiplying through by 2 to clear the fraction, we get:

[tex]\[ 0 = 150 - 40t - 9.8t^2 \][/tex]

Rearranging the terms to align with the standard quadratic form [tex]\( at^2 + bt + c = 0 \)[/tex], we have:

[tex]\[ 9.8t^2 + 40t - 150 = 0 \][/tex]

Now we can apply the Quadratic Formula to solve for t:

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

Where ( a = 9.8 ), ( b = 40 ), and ( c = -150 ). Plugging in these values, we get:

[tex]\[ t = \frac{-40 \pm \sqrt{40^2 - 4(9.8)(-150)}}{2(9.8)} \][/tex]

Solving this, we get two values for t, but only the positive value is physically meaningful since time cannot be negative.

 Calculating the discriminant:

[tex]\[ \sqrt{40^2 - 4(9.8)(-150)} = \sqrt{1600 + 5880} = \sqrt{7480} \][/tex]

Now, we find the two possible values for t:

[tex]\[ t = \frac{-40 \pm \sqrt{7480}}{2(9.8)} \] \[ t = \frac{-40 \pm 86.52}{19.6} \][/tex]

The two solutions for t are:

[tex]\[ t_1 = \frac{-40 + 86.52}{19.6} \approx 2.37 \text{ s} \] \[ t_2 = \frac{-40 - 86.52}{19.6} \approx -6.43 \text{ s} \][/tex]

Since time cannot be negative, we discard [tex]\( t_2 \)[/tex] and take [tex]\( t_1 \)[/tex] as the time when the rock hits the ground.

The time taken is 2.37 s.

Answer:

F = 104.832 N

Explanation:

given,

upward acceleration of the lift = 1.90 m/s²

mass of box containing new computer = 28 kg.

coefficient of friction = 0.32

magnitude of force = ?

box is moving at constant speed hence acceleration will be zero.

Now force acting due to lift moving upward =

               F = μ m ( g + a )

               F = 0.32 × 28 × ( 9.8 + 1.9 )

              F = 104.832 N

hence, the force applied should be equal to 104.832 N

Answer:

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector [tex]\hat{i}[/tex] pointing north and [tex]\hat{j}[/tex] pointing east, the final velocity will be

[tex] \vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )[/tex]

[tex] \vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

The final linear momentum will be:

[tex]\vec{P}_f = (m_{car}+ m_{truck}) * V_f[/tex]

[tex]\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

[tex]\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

[tex]\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]

As there are not external forces, the total linear momentum must be constant.

So:

[tex]\vec{P}_0= \vec{P}_f [/tex]

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

[tex]\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )[/tex] 

so:

 [tex]\vec{P}_0= \vec{P}_f[/tex] 

[tex]( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]  

so

[tex]\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.[/tex]

So, for the truck

[tex]m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]

[tex]1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]

[tex] v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]

[tex]v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]

[tex]v_{truck} = 21.93 \frac{m}{s}[/tex]

And, for the car

[tex]950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}[/tex]

[tex]v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}[/tex]

[tex]v_{car}=19.524 \frac{m}{s}[/tex]

Answer: a) close together

Explanation: The electric field lines also  represent the intensity of the field, in this sense  for strong electric fields it is usually draw the lines close to each other. In constrast when they are far apart the electric field is weak.  

Final answer:

The electric field lines are close together when the electric field strength is large, indicating a stronger field due to the direct proportionality between field strength and the density of the field lines. The correct answer is a) close together.

Explanation:

When the electric field strength is large, electric field lines are close together. This is because the strength of the electric field is directly proportional to the number of lines per unit area perpendicular to the field lines. The more electric field lines there are in a given area, the stronger the electric field at that point.

Hence, near a positive charge or around a negative charge, where the field lines begin or terminate, the field lines are densely packed when the charge is large, indicating a strong electric field. Field lines provide a visual representation of the field's strength and direction, and they will never intersect because the electric field at a point must have a unique direction.

Answer:

the vector position is   r = (10.4 t + t²) i + (8+ 6t - 2 t²) j

Explanation:

In this problem we have acceleration in the x and y axes, therefore we will use the accelerated motion equations

   X = Xo + Vox t + ½ aₓ t²

   Y = I + I go t + ½ [tex]a_{y}[/tex] t²

We calculate the components of velocity using trigonometry

   Vox = Vo cos θ = 12 cos 30

   Voy = Voy sin θ = 12 sin 30

    Vox = 10.4 m / s

    Voy = 6 m / s

The value accelerations are data

    aₓ = 2 m / s²

    [tex]a_{y}[/tex] = - 4 m / s²

The initial position is

    Xo = 0

    Yo = 8 m

    Since the launching point is 8 m above the ground.  The acceleration is negative because it has a downward direction; with this data we write the equations of the position

      X = 10.4 t + ½ 2 t²

      Y = 8 + 6 t - ½ 4 t²

     X = 10.4 t + t²

      Y = 8 + 6 t - 2 t²

We write the position vector

    r = x i + y j

     r = (10.4 t + t²) i + (8+ 6t - 2 t²) j

this the vector position

Answer:

13330.86 J

Explanation:

mass of volume of 30.1 cc of water = density x volume

= 1 x 30.1 = 30.1 gm

= 30.1 x 10⁻³ kg

Heat to be withdrawn to cool water from 22.4 degree to 0 degree

= mass x specific heat x fall of temperature

= 30.1 x 10⁻³ x 4186 x 22.4

= 2822.36 J

Heat to be withdrawn to cool water  0 degree ice

mass x latent heat of freezing

30.1 x 10⁻³ x 334000

= 10053.4 J

Heat to be withdrawn to cool ice from 0 degree to -7.2 degree

mass x specific heat of ice x fall of temperature

= 30.1 x 10⁻³ x 2100 x 7.2

= 455.1 J

Total heat to be withdrawn

=  2822.36  + 10053.4 + 455.1

= 13330.86 J

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.  

Answer:

4.51 * 10^{-5} C

Explanation:

The force between two charge  q and q1 is given as

[tex]F = \frac{k*q*q1}{r^2}[/tex]

[tex]= \frac{(9.0 * 10^9)(24 * 10^{-6})(8 * 10^{-6} C)}{(0.19m)^2} [/tex]

= 47.86 N in the +y direction

We need the force between q and q2 to be (47.86 - 18) =  29.86 N in the other direction to get the desired result.

solving for q2,

[tex]q2 = \frac{Fr^2}{(kq)} [/tex]

[tex]= \frac{(29.86)(0.33 m)^2}{(9.0 * 10^9*8*10^{-6} C)}[/tex]

[tex]= 4.51 * 10^{-5} C[/tex]

Final answer:

To determine the magnitude of q2, we can use Coulomb's Law to calculate the net force exerted on charge q. By setting up an equation using the given information, we can find that the magnitude of q2 is approximately 4.84 μC.

Explanation:

To determine the magnitude of q2, we can use Coulomb's Law, which states that the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, we have two charges, q1 and q2, with a distance of 0.33 - 0.19 = 0.14m between them.

The net electrostatic force on charge q can be calculated using the equation F = k * (|q1 * q| / r1^2) + k * (|q2 * q| / r2^2), where k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2).

From the given information, we know that F = 18 N and points in the +y direction, so we can set up the following equation: 18 = k * (|(-24 μC) * (8 μC)| / (0.19m)^2) + k * (|q2 * (8 μC)| / (0.33m)^2).

Solving this equation, we can find the magnitude of q2, which is approximately 4.84 μC.

Answer:

Explanation:

Y = 5 Sin27( .2x-3t)

= 5 Sin(5.4x - 81 t )

Amplitude = 5 m

Angular frequency ω = 81

frequency = ω / 2π

= 81 / (2 x 3.14 )

=12.89

Wave length λ = 2π / k ,

k = 5.4

λ = 2π / 5.4

= 1.163 m

Phase velocity =ω / k

= 81 / 5.4

15 m / s.

The wave is travelling in + ve x - direction.

Answer:

The horizontal distance is 2.41 mts

Explanation:

For this problem, we will use the formulas of parabolic motion.

[tex]Y=Yo+Voy*t+\frac{1}{2}*a*t^2\\Vx=V*cos(\alpha)\\Vy=V*sin(\alpha)[/tex]

We need to find the time of the whole movement (t), for that we will use the first formula:

we need the initial velocity for that:

[tex]Vy=4.4*sin(24^o)\\Vy=1.79m/s[/tex]

so:

[tex]0=0.70+1.79*t+\frac{1}{2}*(-9.8)*t^2[/tex]

now we have a quadratic function, solving this we obtain two values of time:

t1=0.60sec

t2=-0.234sec

the obvious value is 0.60sec, we cannot use a negative time.

Now we are focusing on finding the horizontal distance.

the movement on X is a constant velocity motion, so:

[tex]x=Vx*t[/tex]

[tex]Vx=4.4*cos(24^o)=4.02m/s\\[/tex]

so:

[tex]x=4.02*(0.60)=2.41m[/tex]

Answer:

Explanation:

Gas law equation for any gas is as follows .

PV = nRT ( for n moles of gas )

Let the pressure , volume , temperature and mole of gas in cylinder A be P,V

T₁ and 2n . Pressure , volume , temperature and mole in cylinder B will be

P , V , T₂ and n.

Applying Gas laws , we get

For gas in cylinder A

PV = 2n R T₁

For gas in cylinder B

PV = n R T₂

Equating these equations , we get

2n R T₁ = n R T₂

2 T₁ = T₂

or,

[tex]2T_A =T_B[/tex]

[tex]T_A[/tex] is less than [tex]T_B[/tex]

Final answer:

Under the conditions of equal pressure and volume, and using the ideal gas law, we determine that the temperatures of the two cylinders of gas must be the same (TA = TB).

Explanation:

Given that the two identical cylinders, A and B, contain the same type of gas at the same pressure, and cylinder A has twice as much gas as cylinder B, we can use the ideal gas law to determine the relationship between their temperatures. The ideal gas law is PV = n ×R× T, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. Since the pressures are equal and the cylinders are identical, volumes are also equal. Thus, because A has twice as much gas as B (nA = 2 x nB), it has twice the number of moles. Holding pressure and volume constant and since R is a constant, both cylinders must have the same temperature for the equation to be balanced. Therefore, we find that TA = TB.

Answer:

The force that acts on the man equals 15354.75 newtons.

Explanation:

After falling through a distance of 24.0 meters the speed of the stunt man upon hitting the mattress can be obtained using third equation of kinematics as

[tex]v^{2}=u^{2}+2gh\\\\\therefore v=\sqrt{2gh}\\\\v=\sqrt{2\times 9.81\times 24}\\\\v=21.7m/s[/tex] ( u=0 since the man falls from rest)

Now since the man is decelerated through a distance of 1.15 meters thus the acceleration produced can be obtained from third equation of kinematics as

[tex]v^{2}=u^{2}+2as\\\\0=u^{2}=2as\\\\a=\frac{-u^{2}}{2s}\\\\a=-\frac{(21.7^{2}}{2\times 1.15}=-204.73m/s^{2}[/tex]

Now by newton's second law the force that produced deceleration of the calculated magnitude is obtained as

[tex]F=mass\times acceleration\\\\F=75.0\times -204.73=-15354.75Newtons[/tex]

The negative sign indicates that the direction of force is opposite of motion.

Answer:Momentum of Truck[tex]=4000\times 20=80,000 kg-m/s[/tex]

Momentum of car[tex]=1350\times 10=13,500 kg-m/s[/tex]

Explanation:

Given

mass of truck(m)=4000 kg

Velocity of Truck is ([tex]V_T[/tex])=[tex]-20\hat{j}[/tex]

mass of car ([tex]m_c[/tex])=1350 kg

Velocity of car[tex](V_c)=10 \hat{j}[/tex]

Conserving momentum

[tex]4000\times \left ( -20\right )+1350\left ( 10\right )=5350v[/tex]

[tex]v=\frac{66,500}{5350}=12.42 m/s[/tex]

Momentum of Truck[tex]=4000\times 20=80,000 kg-m/s[/tex]

Momentum of car[tex]=1350\times 10=13,500 kg-m/s[/tex]

Answer:

1) Momentum of truck before collision is [tex]\overrightarrow{p_{1}}=-80000kgm/s[/tex]

2) Momentum of car before collision is [tex]\overrightarrow{p_{2}}=13500kgm/s[/tex]

Explanation:

The momentum of an object with mass 'm' travelling with speed 'v' is mathematically given by

[tex]\overrightarrow{p}=mass\times \overrightarrow{v}[/tex]

In this problem we shall assume that direction's coincide with the Cartesian axis for simplicity

Thus for Truck we have

Mass = 4000 kg

Velocity = [tex]-20\widehat{j}[/tex]m/s

Thus momentum of truck becomes

[tex]\overrightarrow{p_{1}}=4000\times -20\overrightarrow{j}=-80000kgm/s[/tex]

Similarly for car we have

Mass = 1350 kg

Velocity = [tex]10\widehat{j}[/tex]m/s

Thus momentum of car becomes

[tex]\overrightarrow{p_{2}}=1350\times 10\overrightarrow{j}=13500kgm/s[/tex]