A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.74 s. He breaks the tackle and runs straight forward another 29.0 m in 5.20 s. (a) Calculate his average velocity (in m/s) for each of the three intervals. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
v1= m/s, v2= m/s, v3= m/s
(b) Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
m/s

Final answer:

The work done by the man in pushing the lawn mower is 699.245 J, calculated by determining the horizontal force component and multiplying by the distance pushed.

Explanation:

To calculate how much work is done by the man in pushing the lawn mower, we need to consider only the component of the force that acts in the direction of the movement. Since 37% of the 195 N force is directed downward, only the remaining 63% is contributing to the horizontal movement. Therefore, the horizontal component of the force is 0.63 × 195 N = 122.85 N.

The formula to calculate work (W) is W = force (F) × distance (d) × cosine(θ), where θ is the angle between the force and the direction of movement. In this case, the force and movement are in the same direction, so θ = 0 and cosine(θ) = 1. Thus, the work done is:

W = 122.85 N × 5.7 m × 1 = 699.245 J

In terms of energy expended while pushing a lawn mower, this work is a relatively small amount when compared to a person's daily intake of food energy.

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

[tex]magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km[/tex]

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km  

Answer:

Light travels faster in a less dense medium

Explanation:

In a less dense medium, light will travel faster than it did in the more dense medium (an example of this transition can be light going from water to air).

This is because in a less dense medium there ares less particles, so light finds less resistence in its path.

The opposite happens in a more dense medium, there are more particles so light will travel slower because more particles will get in the way.

In summary, the more dense the medium, the more particles it has, and the slower that light will travel in it.

Answer:

c) Both a) and b) are the combinations that have a negative velocity.

Explanation:

The velocity is given by this equation:

v = Δx / Δt

Where

Δx = final position - initial position

Δt = elapsed time

Now let´s evaluate the options. We have to find those combinations in which  Δx < 0 since Δt is always positive.

a) If the initial position, x, is positive and you move towards the origin, the final position will be a smaller value than x. Then:

                          final position < initial position

final position - initial position < 0 The velocity will be negative.

b) If x is negative and you move away from the origin, the final position will be a more negative number than x. Again:

                          final position < initial position

final position - initial position < 0 The velocity will be negative.

Let´s do an example to show it:

initial position = -5

final position = -10 (since you moved away from the origin)

final position - initial position = -10 -(-5) = -5

d) If x is positive and you move away from the origin, the final position will be a greater value than the initial position. Then:

                          final position > initial position

final position - initial position > 0 The velocity will be positive.

e) If x is negative and you move towards the origin, the final position will be a greater value than the initial position. Then:

                          final position > initial position

final position - initial position > 0 The velocity will be positive.

Let´s do an example:

initial position = -10

final position = -5

final postion - initial position = -5 - (-10) = 5

or with final position = 0

final postion - initial position = 0 -(-10) = 10

And so on.

The right answer is c) Both a) and b) are the combinations that have a negative velocity.

Answer:

a) 94.176 m/s

b) 452.04 m

Explanation:

t = Time taken by the quarter to reach the ground = 9.6 s

u = Initial velocity

v = Final velocity

s = Displacement

a)

[tex]v=u+at\\\Rightarrow v=0+9.81\times 9.6\\\Rightarrow v=94.176 m/s[/tex]

Speed of the quarter just before it hits the ground is 94.176 m/s

b)

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 9.6^2\\\Rightarrow s=452.04\ m[/tex]

The building is 452.04 m high

Answer:

Explanation:

given,

mass of neutron = 1.675 × 10⁻²⁷ kg

1 kg =  5.58 × 10³⁵ eV/c²                                        

a) mass of proton = 1.675 × 10⁻²⁷ ×  5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²

b) mass of electron = 9.109 x 10⁻³¹ × 5.58 × 10³⁵ = 5.08 × 10⁵eV/c²

c) mass of hydrogen = 1.675 × 10⁻²⁷  × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s[/tex]

a) The vertical speed when the player leaves the ground is 4.45 m/s

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s[/tex]

Time taken to reach the maximum height is 0.45 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s[/tex]

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

Answer:

1.05 ms⁻²

Explanation:

Acceleration = change in velocity / Time

Change in velocity = Final velocity - initial velocity

= 1.77 - (-1.29)

= 1.77 + 1.29

= 3.06 m/s

Time = 2.91

Acceleration = 3.06 / 2.91

= 1.05 ms⁻² .

Answer:

3.41 s

114 m

Explanation:

The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.

It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.

0.5*h = 1/2 * a * t1^2

h = a * t1^2

It travels 0.5*h in 1 second.

h = X(t1 + 1) = 1/2 * a * (t1+1)^2

Equating both equations:

a * t1^2 = 1/2 * a * (t1+1)^2

We simplify a and expand the square

t1^2 = 1/2 * (t1^2 + 2*t1 + 1)

t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0

1/2 * t1^2 - t1 - 1/2 = 0

Solving electronically:

t1 = 2.41 s

total time = t1 + 1 = 3.41.

Now

h = a * t1^2

h = 9.81 * 3.41^2 = 114 m

Answer:

31.174°

Explanation:

Bragg's condition is occur when the wavelength of radiation is comparable with the atomic spacing.

So, Bragg's reflection condition for n order is,

[tex]2dsin\theta=n\lambda[/tex]

Here, n is the order of maxima, [tex]\lambda[/tex] is the wavelength of incident radiation, d is the inter planar spacing.

Now according to the question, first order maxima occur at angle of 15°.

Therefore

[tex]2dsin(15^{\circ})=\lambda\\sin(15^{\circ})=\frac{\lambda}{2d}[/tex]

Now for second order maxima, n=2.

[tex]2dsin\theta=2\times \lambda\\sin\theta=\frac{2\lambda}{2d}[/tex]

Put the values from above conditions

[tex]sin\theta=2\times sin(15^{\circ})\\sin\theta=2\times 0.258819045103\\sin\theta=0.517638090206\\\theta=sin^{-1}0.517638090206\\ \theta=31.174^{\circ}[/tex]

Therefore, the second order maxima occurs at 31.174° angle.

Final answer:

The motorist's total displacement is 179.58 km north, and his average velocity is approximately 63.38 km/h north for the entire trip.

Explanation:

Displacement and Average Velocity Calculation

For the given problem, we can find the motorist's total displacement by adding the distances traveled in each segment of the journey, all in the same direction (north). First, the motorist drives north for 35.0 minutes at 85.0 km/h. The distance for this part of the trip is:

Distance = Speed × Time = 85.0 km/h × (35.0 min / 60 min/h) = 49.58 km (rounded to 2 decimal places)

The next segment involves a stop, so the displacement remains unchanged.

Finally, the motorist drives an additional 130 km north.

Therefore, the total displacement is the sum of these distances: Total Displacement = 49.58 km + 130 km = 179.58 km north.

To find the average velocity, we need to consider the total displacement and the total time, including the stop:

Total time = Driving time + Stopping time = (35.0 min + 2.00 h + 15.0 min) = 2.833 hours (2 hours and 50 minutes).

Average Velocity = Total Displacement / Total Time = 179.58 km / 2.833 h ≈ 63.38 km/h north.

The motorist's total displacement is 179.58 km north, and his average velocity over the entire trip is approximately 63.38 km/h north.

Answer:

Final volumen first process [tex]V_{2} = 98,44 cm^{3}[/tex]

Final Pressure second process [tex]P_{3} = 1,317 * 10^{10} Pa[/tex]

Explanation:

Using the Ideal Gases Law yoy have for pressure:

[tex]P_{1} = \frac{n_{1} R T_{1} }{V_{1} }[/tex]

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

[tex]n_{1} = n_{2} = n [/tex]

In an isobaric process the pressure is constant so:

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

[tex]\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }[/tex]

[tex]\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }[/tex]

[tex]V_{2} = \frac{T_{2} V_{1} }{T_{1} }[/tex]

Replacing : [tex]T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}[/tex]

[tex]V_{2} = 98,44 cm^{3}[/tex]

Replacing on the ideal gases formula the pressure at this piont is:

[tex]P_{2} = 3,92 * 10^{9} Pa[/tex]

For Temperature the ideal gases formula is:

[tex]T = \frac{P V }{n R }[/tex]

For the second process you have that [tex]T_{2} = T_{3} [/tex]  So:

[tex]\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }[/tex]

[tex]P_{2} V_{2}  = P_{3} V_{3} [/tex]

[tex]P_{3} = \frac{P_{2} V_{2}}{V_{3}} [/tex]

[tex]P_{3} = 1,317 * 10^{10} Pa[/tex]

Answer:

1.43 x 10¹⁷.

They will accelerate away from each other.

Explanation:

Force on each charged sphere F = mass x acceleration

= 8.55 x 10⁻³ x 25 x 9.8

= 2.095 N

Let Q be the charge on each sphere

F = [tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]

2.095 =[tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]

Q² =[tex]\frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}[/tex]

Q = 2.289 X 10⁻⁶

No of electrons = Charge / charge on a single electron

= [tex]\frac{2.289\times10^{-6}}{1.6\times10^{-19}}[/tex]

=1.43 x 10¹³.

They will accelerate away from each other.

The number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.

Electric force is the force of attraction of repulsion between two bodies.

According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,

[tex]F=\dfrac{KQ_1Q_2}{r^2}[/tex]

Here, (k) is the coulombs constant,  (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.

Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.

Two spheres accelerate at 25.0g when released, and the mass of each sphere is 8.55 g. Thus, the force on the sphere can be given as,

[tex]F=8.55\times{10^{-3}}\times25\times9.8\\F=2.095\rm \;N[/tex]

Two very small spheres are 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.As the charge on both the sphere is same (say Q) Put these values in the above formula as,

[tex]2.095=\dfrac{(9\times10^{9})QQ}{(15\times10^{-2})^2}\\Q=2.289\times10^{-6}\rm \;C[/tex]

It is known that the charge on one electron is 1.6×10⁻¹⁹ C. Thus the number of electron in the above charge is,

[tex]n=\dfrac{2.289\times10^{-6}}{1.6\times10^{-19}}\\n=1.43\times10^{13}[/tex]

The direction of acceleration of both the sphere will be opposite to each other. Thus, they accelerate away from each other.

Hence, the number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.

Learn more about the electric force here;

https://brainly.com/question/14372859

Answer:

Explanation:

Force = mass x acceleration

[tex]\overrightarrow{F}=m\overrightarrow{a}[/tex]

Force is always vector and acceleration also vector but the mass is a saclar quanity.

here, the direction of force vector is same as the direction of acceleration vector but the magnitude of force depends on the magnitude of mass of the body.

Is mass is more, force is also more.

Thus, the mass is like an indicator of the magnitude of force.

Answer:8769 Kg/m^3

Explanation: First of all we have to calculate the volume of the cylinder so it is equal to:

Vcylinder= π*r^2*h where r and h are the radius and heigth, respectively.

Vcylinder= π*0.0275^2*0.048=1.14*10^-4 m^3

The density is defined by ρ=mass/volume

the we have

ρ=1Kg/1.14*10^-4 m^3=8769 Kg/m^3

Answer:

Football will be in air for 2.8139 sec

Explanation:

We have given maximum height h = 9.7 meters

Angle of projection [tex]\Theta =38^{\circ}[/tex]

We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]

So [tex]9.7=\frac{u^2sin^{2}38^{\circ} }{2\times 9.8}[/tex]

[tex]u^2=501.5842[/tex]

u = 22.396 m/sec

Time of flight is given by

[tex]T=\frac{2usin\Theta }{g}=\frac{2\times 22.396\times \times sin38^{\circ}}{9.8}=2.8139sec[/tex]

Answer:

El flujo del campo U a través de dicha superficie es [tex]\Phi_U=4\pi r^2K[/tex]

Explanation:

El flujo vectorial de un campo U a través de una superficie S está dado por

[tex]\Phi_U=\int_S \vec{U} \cdot \hat{n} \ dS[/tex]

donde [tex]\hat{n}[/tex] representa el vector unitario normal a la superficie. Teniendo en cuenta que el campo se encuentra en dirección radial, se tiene

[tex]\Phi_U=\int_S K\hat{r} \cdot \hat{r} \ dS = K\int_S dS = KS=K*4\pi r^2[/tex]

Answer:

a) Acceleration of runner is 1.33 m/s²

b)  Acceleration of motorcycle is 2.85 m/s²

c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2[/tex]

Acceleration of runner is 1.33 m/s²

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2[/tex]

Acceleration of motorcycle is 2.85 m/s²

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m[/tex]

The runner moves 2.94 m

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m[/tex]

The motorcycle moves 84.21 m

The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

Answer:22.62 m/s

Explanation:

Given

two balls are separated by a distance of 50 m

Alex throws  the ball from a height of 50 m with a velocity of 8 m/s and Gary launches a ball with some velocity  exactly at the same time.

ball  from ground travels a distance of 25 m in t sec

For Person on tree  

[tex]25=ut+\frac{1}{2}gt^2[/tex]

[tex]25=8t+\frac{1}{2}\times 9.81\times t^2--------1[/tex]

For person at ground

[tex]23.5=ut-\frac{1}{2}gt^2---------2[/tex]

Solve equation (1)

[tex]50=16t+9.81t^2[/tex]

[tex]9.81t^2+16t-50=0[/tex]

[tex]t=\frac{-16\pm\sqrt{256+4\times 50\times 9.81}}{2\times 9.81}=\frac{47.1-16}{19.62}=1.58 s[/tex]

put the value of t in equation 2

[tex]23.5=u\times 1.58-\frac{9.81\times 1.58^2}{2}[/tex]

[tex]u=\frac{35.744}{1.58}=22.62 m/s[/tex]

Answer:

The minimum time to get the car under max. speed limit of 79 km/h is 2.11 seconds.

Explanation:

[tex]a=\frac{V_f-V_0}{t}[/tex]

isolating "t" from this equation:

[tex]t=\frac{V_f-V_0}{a}[/tex]

Where:

a=[tex]-4.6m/s^2[/tex] (negative because is decelerating)

[tex]V_f= 79 km/h[/tex]

[tex]V_0= 114 km/h[/tex]

First we must convert velocity from km/h to m/s to be consistent with units.

[tex]79\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{79*1000}{3600}=21.94 m/s[/tex]

[tex]114\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{114*1000}{3600}=31.67 m/s[/tex]

So;

[tex]t=\frac{V_f-V_0}{a}=\frac{21.94 m/s-31.66m/s}{-4.6 m/s^2}=2.11 s[/tex]

Answer:

Acceleration of the cart will be [tex]a=2m/sec^2[/tex]

Explanation:

We have given force F = 40 N

Mass of the cart = 20 kg

From newton's second law we know that force, mass and acceleration are related to each other

From second law of motion force on any object moving with acceleration a is given by

F = ma, here m is mass and a is acceleration

So [tex]40=20\times a[/tex]

[tex]a=2m/sec^2[/tex]

Answer:

The inicial height of the apple is 1.22 meters

Explanation:

Using the equation for conservarion of mechanical energy:

[tex]E=V+K=constant[/tex]

[tex]K_i=\frac{1}{2}mv_i^2[/tex] where v is the velocity

[tex]V=mgh[/tex]where h is the height

We equate the initial mechanical energy to the final:

Since [tex]v_0=0\ and h_f=0 [/tex]:

[tex]\frac{1}{2}mv_0^2+mgh_0= \frac{1}{2}mv_f^2+mgh_f\\gh_0= \frac{1}{2}v_f^2[/tex]

Solving for h:

[tex]h_0=\frac{4.9^2}{2g}= 1.22 m[/tex]

Answer:

Clockwise, converging towards

Explanation:

At the center of surface , system with low pressure in Northern hemisphere have air around the center that rotates in anti clockwise direction.

As a result of low pressure, the air is directed slightly inwards thus converges towards the center of the system.

In the high pressure systems, air rotates in clockwise direction and diverges away from the center of the system.

Answer:

T = 61.06 °C  

Explanation:

given data:

a thin metal bar consist of 5 different material.

thermal conductivity of ---

K {steel} = 16 Wm^{-1} k^{-1}

K brass = 125 Wm^{-1} k^{-1}

K copper = 401 Wm^{-1} k^{-1}

K aluminium =30Wm^{-1} k^{-1}

K silver = 427 Wm^{-1} k^{-1}

[tex]\frac{d\theta}{dt} = \frac{KA (T_2 -T_1)}{L}[/tex]

WE KNOW THAT

[tex]\frac{l}{KA} = thermal\ resistance[/tex]

total resistance of bar = R steel + R brass + R copper + R aluminium + R silver

[tex]R_{total} =\frac{1}[A} [\frac{0.02}{16} +\frac{0.03}{125} +\frac{0.01}{401} +\frac{0.05}{30} +\frac{0.01}{427}][/tex]

[tex]R_{total} =\frac{1}[A} * 0.00321[/tex]

let T is the temperature at steel/brass interference

[tex]\frac{d\theta}{dt}[/tex] will be constant throughtout the bar

therefore we have

[tex]\frac{100-0}{R_{total}} = \frac{100-T}{R_{steel}}[/tex]  

[tex]\frac{100-0}{0.00321} *A = \frac{100-T}{0.00125} *A[/tex]

solving for T  we get

T = 61.06 °C  

Answer:

The right answer is (D) 340m

Explanation:

If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.

The initial speed of the package V₀ is 15m/s.

The acceleration of the package G is 9.8m/s².

The initial hight is Y₀. Therefore:

Y(t)=Y₀+V₀·t-0.5·G·t²

But we know than T(10s)=0m

0m=Y₀+150m-490m

Y₀=340m

Answer:

Not the right answer in the options, speed is 4.47 m/s, and the procedure is coherent with option A

Explanation:

Answer A uses mass and velocity units, which are momentum units. By using the conservation of momentum:

.[tex]p_{initial} =p_{final} \\m_{Tom}*v_{Tom}+m_{raft}*v_{raft}=(m_{Tom}+m_{raft})*v_{both} \\70*10+130*1.5 kg*m/s=895kg*m/s\\v_{both}=\frac{895 kg*m/s}{200 kg} =4.47 m/s[/tex]

Since Tom stays in the raft, then both are moving with the same speed. From the options, the momentum is in agreement with option A, however, the question asks for speed.

Answer:

Focal length of the lens, f = - 68 mm

Explanation:

Given that,

Power of a lens, P = -14.50 D

We need to find the focal length of the lens. We know that the focal length and the power of lens has inverse relationship. Mathematically, it is given by :

[tex]f=\dfrac{1}{P}[/tex]

f is the focal length of the lens

[tex]f=\dfrac{1}{-14.50}[/tex]

f = -0.068 m

or

f = -68 mm

So, the focal length of the lens is (-68 mm). Hence, this is the required solution.

Answer:

After 0.19288 sec second object hit the ground after hitting first object    

Explanation:

We have given two objects are dropped from at the same time from height 0.49 m and 1.27 m

As the object is dropped so its initial velocity will be zero u = 0 m/sec

For object 1 height h = 0.49 m

From second equation of motion

[tex]h=ut+\frac{1}{2}gt^2[/tex]

[tex]0.49=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]

[tex]t^2=0.1[/tex]

t = 0.31622 sec

For second object

[tex]h=ut+\frac{1}{2}gt^2[/tex]

[tex]1.27=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]

[tex]t^2=0.1[/tex]

t = 0.5091 sec

So difference in time = 0.5091-0.31622 = 0.19288 sec

So after 0.19288 sec second object hit the ground after hitting first object

Answer:

absolute pressure =  1.07 lbft/in^2

Explanation:

given data:

vaccum gauge reading h = 27.86 inch = 2.32 ft

we know that

gauge pressure p is given as

[tex]p = \rho gh[/tex]

[tex]p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft  = 62955.52 \ lbft/s^2 * 1/ft^2[/tex]

we know that  [tex]1\  lb ft\s^2 = \frac{1}{32.174}\  lbft[/tex]

1 ft = 12 inch

therefore [tex]p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2[/tex]

               [tex]p = 13.59\  lbft/in^2[/tex]

[tex]P_{atm} = 14.66\  lbf/in^2[/tex]

so absolute pressure = [tex]P_{atm} - p[/tex]

                                   = 14.66 - 13.59 = 1.07 lbft/in^2

Answer:595.32 N

Explanation:

Given

Mechanic shop has a car of weight([tex]F_1[/tex]) 16,500 N

Area of lift([tex]A_1[/tex])=[tex]125 cm^2[/tex]

Area of small piston([tex]A_2[/tex])=[tex]4.51 cm^2[/tex]

According to pascal's law pressure transmit will be same in all direction in a closed container

[tex]P_1=P_2[/tex]

[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

[tex]F_2=F_1\frac{A_2}{A_1}[/tex]

[tex]F_2=16,500\times \frac{4.51}{125}=595.32 N[/tex]