The box now rests on a frictionless ramp angled at 15◦ . The mover pulls up on a rope attached to the box to move it up the incline. 33 kg F 29 ◦ 15◦ If the rope makes an angle of 29 ◦ with the horizontal, what is the smallest force the mover would have to exert to move the box up the ramp? The acceleration due to gravity is 9.81 m/s 2 .

Answer:

Time, t = 2 minutes

Explanation:

Given that,

Length of Duplain st. d = 300 m = 0.3 km

If keith is strolling east from Baron at an average velocity of, v = 3 km/hr

Sue is power-walking west from Burkey at an average velocity of, v' =  6 km/hr

To find,

How long will it take them to meet ?

Solution,

When both objects are travelling in opposite direction, then the total speed is given by :

V = v + v'

V = 3 km/hr + 6 km/hr

V = 9 km/hr

Let t is the time taken will it take them to meet. It can be calculated as :

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

[tex]t=\dfrac{0.3\ km}{9\ km/hr}[/tex]

t = 0.033 hour

or

t = 1.98 minutes

i.e. t = 2 minutes

So, they will take 2 minutes to meet. Hence, this is the required solution.

Answer:

d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion

Explanation:

The speeds of the particles are very large and comparatively the average  free path is very small . Therefore time taken in covering the free path ( path between two consecutive collision with medium particles ) is very small . Hence the st line  path covered by particles between two collision is less likely to be visible. Hence motion appears irregular or zig-zag.

Answer:

(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength

Explanation:

de Broglie wavelength λ = h / m v

Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .

for electron , momentum is less so  . Therefore de Broglie wavelength λ will be more for electrons .

Amount of diffraction that is angle of diffraction is proportional to λ

Therefore electrons having greater de Broglie wavelength will show greater diffraction.

Answer:

Rate of plane in still air = P = W (t1 +t2)/ (t1-t2)

Rate of wind in still air = W = P (t1 - t2)/(t1 + t2)

Explanation:

Assuming speed of plane are the same on both trips

Rate (D/t1) = (P-W).... EQU 1 going from city a to b

Rate (D/t2) = (P +W)...Equ2 going back to city a

Where t1 is not equal to t2

Where D=distance between two cities

P &W are the speed of plane and wind

t1 &t2 = time taken for travel

Equ 1 & equ 2 becomes

D = ( P - W ) t1.. equ3

D = ( P + W ) t2 equa4

Equating equ 3 and 4

Pt1 - Wt1 = Pt2 +Wt2

P ( t1 - t2) = W ( t1 + t2)

Rate of plane in still air = P = W (t1 +t2)/ (t1- t2)

Rate of wind in still air = W = P (t1 - t2)/(t1 + t2)

Answer:

[tex]Q=35011.6\ J[/tex]

Explanation:

Given:

Now, total heat energy required get to the final state:

[tex]Q=m(c_i.\Delta T_i+L+c_w.\Delta T_w)[/tex]

where:

[tex]\Delta T_w=[/tex] change in temperature of water from 0 to 73 degree C

[tex]\Delta T_i=[/tex] change in temperature of ice from -30 to 0 degree C

[tex]\therefore Q=50(2.06\times 30+333+4.184\times 73)[/tex]

[tex]Q=35011.6\ J[/tex]

Answer:

The heat energy required is 350.1J

Explanation:

Given data

Mass of ice =50g---kg =50/1000= 0.05kg

Temperature of ice T1= - 30°c

Temperature of ice T2=0°c

Temperature of liquid T3=0°c

Temperature of water T4= 73°c

The heat of fusion = 333 J/g

heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK

and liquid water = 4.184 J/gK

It will be a good idea to first understand the path this process will follow

Ice at - 30°c - - ice at 0°c

Ice at 0°c - - - - liquid at 0°c fusion

Liquid at 0°c -- - liquid at 73°c

First, you have to calculate the heat absorbed by ice going from -30 C to 0 C. Use the equation:

q = m c (T2-T1)

q= 0.5*2.06(0-(-30))

q= 30.9J

Then, calculate the heat required to melt that ice at 0C. Use the equation:

q = m *(heat of fusion)

q=0.5*333

q=166.5J

Then, calculate the heat required to raise the temperature of water from 0°C to 73°C. Use

q = m c (T4-T3)

q= 0.5*4.184(73-0)

q= 152.7J

Finally, we will sum up the heat required

Total heat energy required = 30.9+166.5+152.7= 350.1J

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

[tex]\displaystyle{t=D/v}[/tex]

and from Hubble's Law:

[tex]v =H_0D[/tex]

Therefore:

[tex]\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}[/tex]

With the given value for the Hubble's constant we have:

[tex]H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s[/tex]

and thus,

[tex]t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years[/tex]

Answer:

The work required to move a planet's satellite is W = 2854.61 J

Explanation:

Given data,

The mass of the satellite, m = 1820 kg

The radius of the circular orbit, r =  2R

The radius of the planet, r = 5.37 x 10⁶ m

The mass of the planet, M = 7.5 x 10²⁴ kg

The formula for work done from the 2R to 3R is,

                            W = [tex]\int_{2R}^{3R}\frac{GMm}{r^{2}}dr[/tex]

                            W = GMm/3R - GMm/2R

                             W = (-0.17)GMm/R  J

The negative sign indicates that the energy stored in the satellite as the potential energy.

Substituting the values

                              W = (-0.17) 6.673 x 10⁻¹¹ X 7.51 x 10²⁴ X 1820 / (7.37 x 10⁶)²

                                  = -2854.61 J

Hence, the work required to move a planet's satellite is W = 2854.61 J

Answer:

Angular displacement of the wheel, [tex]\theta=267.41\ rad[/tex]

Explanation:

It is given that,

Angular acceleration of the wheel, [tex]\alpha =6.7\ rad/s^2[/tex]

Final speed of the wheel, [tex]\omega_f=74.5\ rad/s[/tex]

Time taken, t = 4.5 s

Initially, it is required to find the initial angular velocity of the wheel. Using the first equation of rotational kinematics as :

[tex]\omega_f=\omega_o+\alpha t[/tex]

[tex]\omega_o[/tex] is the initial speed of the wheel

[tex]\omega_o=\omega_f-\alpha t[/tex]

[tex]\omega_o=74.5-6.7\times 4.5[/tex]

[tex]\omega_o=44.35\ rad/s[/tex]

Let [tex]\theta[/tex] is the angular displacement of each wheel during this time. Using the second equation of motion as :

[tex]\theta=\omega_o t+\dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta=44.35\times 4.5+\dfrac{1}{2}\times 6.7\times (4.5)^2[/tex]

[tex]\theta=267.41\ rad[/tex]

So, the angular displacement of each wheel during this time is 267 radian.

Answer:

Gold nugget diameter    22,24 cm

Silver nugget price   38860  $

Explanation:  

Gold price    1400  $ /ou            1  ounce  = 31.1035 grs

so  1400 /31.1035

Gold price is 45,01   $/grs.

If a nugget worth   5000000 $  then  5000000/ 45.01

a nugget mass  :  111086,43 grs

Now gold density  is  d = 19.3 grs/cm³

And she volume is  V = 4/*π*r³

d = m/V         V = m/d           V  =   111086.43/19.3    cm³

V = 5755,77 cm³

Now V of the sphere is  V =  5755,77 = 4/3*π*r³

r³  =3*5755,77 / 4π        r³  = 1374,79

r  =  11.12 cm      2r  =  22,24  =  Φ (sphere diameter)

B)  d (silver)  =  m/V       m = d*V

V = 5755,77 cm³     The same size the same volume

m =  10,5 * 5755,77 [grs/cm³ * cm³]       m  = 60435,59 grs

Silver nugget worth  :

20  $ /ou         20/31.1035  =   0.643  $ /grs

Price   0,643 *  60435,59  =  38860  $

Answer:

so throwers new velocity = 2.18032m/s

so catchers new velocity = 0.02577m/s

Explanation:

Directly by conservation of momentum we can write

[tex]m_1u_1+m_2u_2= m_1v_1+m_2v_2[/tex]

let x be the thrower's new velocity

(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0

x = 2.18032m/s

so the velocity of 70 kg man = 2.18032m/s

so throwers new velocity = 2.18032m/s

now again by conservation of momentum

0.042×35 = (57+0.042) ×y

y = 0.02577m/s

so catchers new velocity = 0.02577m/s

Answer:

Running or Jogging

Running and jogging are both great options for aerobic conditioning. Whether you run at the gym or outside, you are in control of setting the intensity of your workout. When aiming to build muscle mass, you can add more resistance or jog at an incline, along with increasing your speed.

Explanation:

Answer:

Running

Jogging

Yoga

Explanation:

Running Jogging and yoga best describes aerobic exercise plan. Aerobic exercises are those where cardiovascular conditioning happen it can be swimming, cycling ,running, jogging etc. aerobic exercise is good for health as it reduces risk of:- obesity, high blood pressure type 2, diabetes, metabolic syndrome , strokes and some type of cancer too can be avoided by doing aerobic exercise.

Answer:

Changes the element of the atom.

Explanation:

The elements are classified by the number of protons they have in their nucleus, so if the number of protons is changed (added or removed), that atom will become one of a different element.

For example, hydrogen has only 1 proton in its nucleus, and helium has 2. So if a proton is added to a hydrogen atom, it becomes a helium atom, and consequently its atomic number, wich is determided by the protons in an element, will also change.

Final answer:

The linear velocity of the space telescope is about 17,193.2 miles per hour when rounded to the nearest tenth. This is calculated using the orbit radius of 4340 miles and the orbital period of 1.5833 hours.

Explanation:

To calculate the linear velocity of a space telescope traveling around Earth, we need to use the formula v = 2πr / T, where v is the linear velocity, r is the radius of the orbit, and T is the orbital period. To find the orbit radius, we add Earth's radius and the altitude of the telescope above Earth's surface: r = 3960 miles + 380 miles = 4340 miles.

Next, we convert the orbital period from minutes to hours. There are 60 minutes in an hour, so T = 95 minutes / 60 minutes/hour = 1.5833 hours. We can then plug the radius and period into the formula to find the linear velocity: v = 2 x π x 4340 miles / 1.5833 hours = 17,193 miles/hour. When rounded to the nearest tenth, the linear velocity is approximately 17,193.2 miles per hour.

The frequency of the yellow light is [tex]5.17\cdot 10^{14}Hz[/tex]

Explanation:

The relationship between wavelength, frequency and speed of a wave is given by the wave equation:

[tex]v = f \lambda[/tex]

where

v is the speed of the wave

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the yellow light in this problem, we have:

[tex]v=1.97\cdot 10^8 m/s[/tex] is the speed

[tex]\lambda=3.81\cdot 10^{-7} m[/tex] is the wavelength

Solving for f, we find its frequency:

[tex]f=\frac{v}{\lambda}=\frac{1.97\cdot 10^8}{3.81\cdot 10^{-7}}=5.17\cdot 10^{14}Hz[/tex]

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The frequency of the yellow light in the glass block is [tex]5.17 \times 10^{14}Hz[/tex]

Yellow light travels through a certain glass block at a speed of 1.97 × 108 m/s. The wavelength of the light in this particular type of glass is 3.81 × 10−7 m (381 nm).

[tex]= 1.97 \times 10^8 \div 3.81 \times 10^{-7}\\\\= 5.17 \times 10^{14}Hz[/tex]

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Answer: 0.58

Explanation:

First we need to get the acceleration of the body using equation of motion

v²=u²-2as

v is the final velocity

u is the initial velocity

a is acceleration

s is the distance moved

0²=9²-2a(2.6)

-81=-5.2a

a=81/5.2

a= 15.6m/s²

Angle of inclination =30°

To get the coefficient of friction, we use the formula

Ff =nR

Ff is frictional force

n is coefficient of friction

R is normal reaction

n = Ff/R = Wsin30°/Wcos30°

n = tan30°

n = 0.58

Answer:

i) C decreases

ii) Q remains constant

iii) E remains constant

iv) ΔV increases

Explanation:

i)

We know, capacitance is given by:

[tex]C=\frac{\epsilon_0.A}{d}[/tex]

[tex]\therefore C\propto \frac{1}{d}[/tex]

In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.

ii)

As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.

iii)

Electric field between the plates is given as:

[tex]E=\frac{\sigma}{\epsilon_0}[/tex]

where:

charge density, [tex]\sigma=\frac{Q}{A}[/tex]

As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.

iv)

i) Capacitance is decreases

ii) Charge Q remains constant

iii) Electric field E remains constant

iv) Change in potential ΔV is  increases

The capacitance is computed as,

                    [tex]C=\frac{\epsilon A}{d}[/tex]

Where A is area of plates and d is distance between plates.

Following information is to be considered.

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Answer:

Angular speed will reach 6.833rad/s before the coin starts slipping

Explanation:

There is no question but I'll asume the common one: Calculate the speed of the turntable before the coin starts slipping.

With a sum of forces:

[tex]Ff = m*a[/tex]

[tex]Ff=m*V^2/R[/tex]

At this point, friction force is maximum, so:

[tex]\mu*N=m*V^2/R[/tex]

[tex]\mu*m*g=m*V^2/R[/tex]

Solving for V:

[tex]V=\sqrt{\mu*g*R}[/tex]

V=1.025 m/s

The angular speed of the turntable will be:

ω = V/R = 6.833 rad/s   This is the maximum speed it can reach before the coin starts slipping.

Final answer:

The photoelectric effect supports the particle theory of light as it shows that light's ability to eject electrons depends on its frequency, indicative of light's particle-like behavior through photons.

Explanation:

The photoelectric effect does not support the wave theory of light, but rather supports the particle theory of light. According to classical wave theory, energy of light is related to its intensity or amplitude. However, the photoelectric effect demonstrates that the kinetic energy of ejected electrons from a metal surface depends on the light's frequency rather than its intensity.

This phenomenon can be explained by considering light to consist of particles called photons. Each photon carries a quantized amount of energy determined by the equation E = hv, where E is energy, h is Planck's constant, and v is frequency. If the frequency of light is above a certain threshold, it can dislodge electrons from the metal because the photons have sufficient energy, showing light's particle-like nature.

Answer:

a) The screen should be located at 1.08 meters

b) The width of the central maximum is 1.7 mm

c) See figure below.

Explanation:

a) This is a single slit diffraction problem, the equation that describes this kind of phenomenon is:

[tex]a\sin\theta=m\lambda [/tex] (1)

Because we’re interested in a minimum near the center of the screen, we can use the approximation [tex] \sin\theta\approx\tan\theta=\frac{y}{x} [/tex]

So equation (1) is now:

[tex]a\frac{y}{x}=m\lambda [/tex] (2)

Solving (2) for x:

[tex] x=\frac{ay}{m\lambda}=\frac{(0.75\times10^{-3})(0.85\times10^{-3})}{1(587.5\times10^{-9})}\approx1.08m [/tex]

b) As you can see on the figure below a maximum is approximately between the two adjacent minimums, because the diffraction pattern is approximately symmetric respect the center of the screen the width of the central maximum is 2*0.850mm = 1.7 mm.

Answer:

32 bottles

Explanation:

If we create a free body diagram on the child we have his weight and the bouyant force

W-B=0

They must be equal to mantain equilibrium on the body and he can stay floating, this force is equivalent to the weight of water displaced

W=B=Ww

Mg=mg

32 kg=mass of water displaced

1 kilogram per liter (kg/L) is the density of water, this means that 32 Liters of water are displaced and since the bottles can retain 1 liter, the child needs 32 bottles

The child needs a minimum of 32 soda bottles to stay dry on the raft.

To determine the minimum number of soda bottles needed for the child to stay dry on the raft, we need to consider the buoyant force exerted by the bottles. The buoyant force is equal to the weight of the displaced water. Since the child wants to stay dry, the buoyant force should be greater than or equal to the weight of the child.

The weight of the child can be calculated using the formula: weight = mass × gravity. Given the mass of the child is 32 kg, and the acceleration due to gravity is 9.8 m/s², we can find that the weight of the child is 32 kg × 9.8 m/s² = 313.6 N.
Next, we need to find the volume of one soda bottle. Since the empty soda bottles have a total volume of 1.0 L, we can assume that each bottle has a volume of 1.0 L ÷ the number of bottles needed. The mass of the water displaced by one bottle can be calculated using the formula: mass = density × volume. Given that the density of water is 1000 kg/m³, and 1 L = 0.001 m³, we can find that the mass of water displaced by one bottle is 1000 kg/m³ × 0.001 m³ = 1 kg.

To find the minimum number of bottles needed, we can set up the equation: buoyant force = weight of child. The buoyant force is equal to the mass of water displaced by one bottle × gravity × the number of bottles needed. Using the values we found earlier, we have: 1 kg × 9.8 m/s² × the number of bottles needed = 313.6 N. Solving for the number of bottles needed, we find that the minimum number of soda bottles the child needs is 313.6 N / (1 kg × 9.8 m/s²) = 32 bottles (rounded up to the nearest whole number).

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If the atoms that share electrons have an unequal attraction for electrons, the bond is called a Polar covalent bond.

A covalent chemical bond is formed in case of two different non-metals when one or more electron pairs are shared between bonding atoms. A difference in electronegativity of subsequent atoms of a covalent bond leads to formation of a small net charge around nucleus of each atom, pulling the shared electrons to one side of the bond, to the nucleus which has higher electronegativity.

HCl is an example of polar covalent bond and the HCl bond has Chlorine more electronegative. The bonding electrons are more close to Cl than H and hence Cl is partially negatively charged than H which has partial positive charge (HCl bond : [tex]H^{+} - Cl^{-}[/tex]). When electrons shared in a covalent bond have equal attraction, the bond is a Non-Polar covalent bond.

The velocity of the ball is 7 m/s

Explanation:

The motion of the ball is a free fall motion, so it means that the ball falls down under the effect of the force of gravity only. Therefore, it has a constant acceleration (acceleration of gravity, g), and we can use the following suvat equation:

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

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the ball in this problem, we have:

u = 0 (initial velocity, the ball is dropped from rest)

[tex]a=g=9.8 m/s^2[/tex] (acceleration of gravity)

s = 2.5 m (vertical displacement)

Solving for v, we find the velocity at which the ball hits the concrete surface:

[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(2.5)}=7 m/s[/tex]

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Answer:

B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.

Explanation:

The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:

• They live at least a few billion years, allowing life a chance to evolve. More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.

• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.

• They emit sufficient radiation at wavelengths conducive to photosynthesis.

• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.

Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.

Answer:

x = 7.62 m

Explanation:

First we need to calculate the weight of the rocket:

W  = mg

we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:  

W  = 0.5 * 9.8 = 4.9 N

We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:

F - W = m*a

a = F - W/m

Solving for a:

a = (8 - 4.9) / 0.5

a = 6.2 m/s²

As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:            

y = vo*t + 1/2 at²

y = 1/2at²

Solving for t:

t = √2y/a

t = √2 * 20 / 6.2

t = √6.45 = 2.54 s

Now that we have the time, we can calculate the horizontal distance:

x = V*t

Solving for x:

x = 3 * 2.54 = 7.62 m            

Answer:

a)1.385  rad/s

b) Before: 1080 J. After 498.46 J

Explanation:

The moments of inertia of the turn table, with the shape of uniform disk is:

[tex] I_1 = 0.5mr^2 = 0.5*120*2^2 = 240 kgm^2[/tex]

The angular momentum of the turn table before the impact is

[tex]A_1 = \omega_1I_1 = 3*240 = 720 radkgm^2[/tex]

The moments of inertia of the system after the impact is (treating the parachute man is a point particle)

[tex]I_2 = I_1 + Mr^2 = 240 + 70*2^2 = 240 + 280 = 520 kgm^2[/tex]

According to angular momentum conservation law:

[tex]A_1 = A_2[/tex]

[tex] 720  = \omega_2I_2[/tex]

[tex]\omega_2 = \frac{720}{I_2} = \frac{720}{520} = 1.385 rad/s[/tex]

(b) Before the impact:

[tex]K_1 = 0.5*I_1*\omega_1^2 = 0.5*240*3^2 = 1080 J[/tex]

After the impct

[tex]K_2 = 0.5*I_2*\omega_2^2 = 0.5*520*1.385^2 = 498.46 J[/tex]

The kinetic energies are not equal because the impact is causing the turn table to lose energy.

Answer:

C. Yes, the water could be changing the phase.

Explanation:

Answer:

1.0x10^-4

Explanation:

First, in order to do this, we need to calculate the volume of 1 simple atom of Ar. Using the formula of the volume of a sphere we have the following

Converting A to cm:

0.97 * 1x10^-8 = 9.7x10^-9 cm

Now the volume:

V = 4/3π(9.7x10^-9)³

V = 3.82x10^-24 cm³

We know that 1 cm³ is 1 mL, and 1 L is 1000 mL so:

V = 3.82x10^-24 mL / 1000 = 3.82x10^-27 L

Now, using avogadro's number, we should get the total volume of all atoms of Ar so:

3.82x10^-27 * 6.02x10^23 = 2.3x10^-3 L

Finally, at STP the volume of an ideal gas is 22.4 L so:

2.3x10^-3 / 22.4 = 1.03x10^-4

With two significant figure, it would be 1.0x10^-4

Answer:

Final velocity will be -5.33 m/sec

Explanation:

We  have given mass of the rugby player m = 106 kg

Initial speed = 7.75 m/sec

Backward force [tex]F=1.85\times 10^4N[/tex]

Time is given as [tex]t=7.5\times 10^{-2}sec[/tex]

Impulse is given by impulse = force × time

So impulse [tex]=-1.85\times 10^4\times 7.5\times 10^{-2}=-1387.5N-s[/tex] ( as force is backward )

We know that impulse is given by change in momentum

So [tex]m(v_f-v_i)=-1387.5[/tex]

[tex]106\times (v_f-7.75)=-1387.5[/tex]

[tex]v_f=-5.33m/sec[/tex]

Answer:

- Elevated temperatures.

- Elevated pressures.

Explanation:

In the manufacture of plastic products are used:

Raw material (pellets) which are the monomers that promote the chemical reaction.

To these are added the charges, in order to reduce the cost of the final product and improve some properties. These charges can be fiberglass, paper, metal structures.

Additives are also added whose mission is to improve or achieve certain properties, such as reducing friction, reducing chemical degradation, increasing electrical conductivity, coloring the product, and all this happens in the presence of a catalyst that is responsible for initiating and accelerating the chemical reaction process.

There are different methods of production of plastics, one of the most frequent is by injection.

Injection is a process that is carried out on machines similar to extrusion machines, in which the spindle, in addition to rotating, has an axial displacement.

In the injection, once the mold is filled, it is separated from the nozzle of the machine, breaking the feed channel. After a certain time, the piece already cooled is demoulded.

High pressures and temperatures are necessary, but parts of good finish and at high production speeds are obtained

The correct answer is: Elevated temperatures.

During fabrication processes, polymeric materials are often subjected to elevated temperatures. This is because many polymers are thermoplastic, meaning they soften and can be reshaped when heated. The elevated temperatures allow the polymer chains to move more freely, enabling the material to be molded, extruded, or otherwise formed into the desired shape or structure.

While ambient temperature and pressure can also affect polymeric materials, the key condition for shaping and fabricating these materials is typically elevated temperature. Ambient temperature is usually not sufficient to induce the necessary molecular mobility for fabrication. Ambient pressure is generally constant and does not play a significant role in the fabrication process unless specific pressure conditions are required for certain molding techniques, such as injection molding or compression molding, where controlled pressure is applied to ensure the polymer fills the mold properly.

Answer:

Part a)

[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]

Part b)

[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]

Part c)

[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]

Explanation:

As per Newton's II law we know that

F = ma

so we will have

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

so we will have

[tex]a = \frac{F_1 + F_2}{m}[/tex]

Part a)

[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}[/tex]

[tex]a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}[/tex]

[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]

Part b)

[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}[/tex]

[tex]a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}[/tex]

[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]

Part c)

[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}[/tex]

[tex]a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}[/tex]

[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]