A ball is thrown straight up, reaches a maximum height, then falls to its initial height. Make a statement about the direction of the velocity and acceleration as the ball is coming down.

Answer:

[tex]MA=\frac{F_l}{F_e}>1[/tex]

Explanation:

The mechanical advantage with a shorter lever will always be greater than 1.

It is so because with a longer effort arm we need to apply lesser force to lift a unit mass which is at a shorter distance from the fulcrum. This is in accordance with the conservation of moments.

[tex]F_e\times d_e=F_l\times d_l[/tex]

where:

[tex]F_e=[/tex] force on the effort arm

[tex]F_l=[/tex] force on the load arm

[tex]d_e\ \&\ d_l[/tex] are the lengths of load effort arm and load arm respectively.

So, one factor balances the other keeping the product of the two constant.

And we know that mechanical advantage is :

[tex]MA=\frac{F_l}{F_e}[/tex]

Answer:

a) 31.02 m/s

b) - 653.24 m/s²

c) 50 ms

Explanation:

a) For a vertical movement, let's suppose that the woman fell by the rest, thus she's initial velocity v0 = 0 m/s. If she makes a distance (S) of 161 ft = 49.1 m, thus, her final velocity at the box, v is:

v² = v0² +2aS

The acceleration is the gravity acceleration, 9.8 m/s².

v² = 2*9.8*49.1

v² = 962.36

v =√962.36

v = 31.02 m/s

b) She crushes the box until she stops, so her final velocity will be 0 m/s. The initial velocity is now 31.02 m/s, and S = 29 in = 0.7366 m

0 = 31.02² + 2a*0.7366

-1.4732a = 962.36

a = - 653.24 m/s² (the negative signal indicates that she's dessacelaranting)

c) The time can be calculate by:

v = v0 + at

0 = 31.02 -653.24*t

653.24t = 31.02

t = 0.05 s = 50 ms

Answer:

The new electric field strength of this charge is half of the strength of the other charge.

Explanation:

The electric force acting on the charge particle is given by :

F = q E

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

Where

q is the charged particle

E is the electric field

If the magnitude of a charge is twice as much as another charge, q' = 2q, but the force experienced is the same, then the new electric field is given by :

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

[tex]E'=\dfrac{F}{(2q)}[/tex]

[tex]E'=\dfrac{1}{2}\times \dfrac{F}{q}[/tex]

[tex]E'=\dfrac{E}{2}[/tex]

So, the new electric field strength of this charge is half of the strength of the other charge. Hence, this is the required solution.

Final answer:

The electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge because the force experienced by both is equal.

Explanation:

If the magnitude of a charge is twice as much as another charge, but the force experienced by each is the same, then the electric field strength that the larger charge is in is half the strength of the other charge’s electric field. The electric field strength (E) at a point is defined as the force (F) experienced by a positive test charge (q) placed at that point divided by the magnitude of the charge itself: E = F/q.

If we have two charges, q1 and q2, where q2 is twice q1 (“q2 = 2 × q1”), and the forces on both charges are equal (“F1 = F2”), we can set up the following equations: E1 = F1/q1 and E2 = F2/q2. Combining our known relationships, we get E1 = F/q1 and E2 = F/(2 × q1), which simplifies to E1 = 2 × E2. Therefore, E2 is half the magnitude of E1, meaning that the electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge.

Answer:

Explanation:

There are two prevailing theories: One is that the Earth held onto some water when it formed, as there would have been ice in the nebula of gas and dust (called the proto-solar nebula) that eventually formed the sun and the planets about 4.5 billion years ago. Some of that water has remained with the Earth, and might be recycled through the planet's mantle layer, according to one theory.

The second theory holds that the Earth, Venus, Mars and Mercury would have been close enough to that proto-solar nebula that most of their water would have been vaporized by heat; these planets would have formed with little water in their rocks. In Earth's case, even more water would have been vaporized when the collision that formed the moon happened. In this scenario, instead of being home-grown, the oceans would have been delivered by ice-rich asteroids, called carbonaceous chondrites.

Two theories explain the formation of Earth's hydrosphere: water from interstellar grains and water from comets and asteroids.

There are two competing scientific theories supported by evidence that explain the formation of Earth's hydrosphere:

Water from comets and asteroids: Another theory suggests that the water may have been brought to Earth when comets and asteroids impacted it. Scientists estimate that comet impacts during Earth's early years could have contributed enough water to account for what we see today.

The linear velocity of the 100 g ball in the rotating system with a 130 g ball and a 230 g ball, connected by a rigid rod and rotating at 120 RPM, is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]

Given values:

- Mass of the 130 g ball [tex](\(m_1\))[/tex]: 130 g

- Mass of the 230 g ball [tex](\(m_2\))[/tex]: 230 g

- Length of the rod [tex](\(r\)):[/tex] 34 cm = 0.34 m

- Initial angular velocity [tex](\(ω_{\text{initial}}\))[/tex]: [tex]\( ω_{\text{initial}} = \frac{2π \times 120}{60} \)[/tex]

Now, let's calculate the initial angular velocity:

[tex]\[ ω_{\text{initial}} = \frac{2π \times 120}{60} = 4π \, \text{rad/s} \][/tex]

Next, calculate the moment of inertia for each ball:

[tex]\[ I_1 = \frac{2}{5}m_1r^2 = \frac{2}{5} \times 0.13 \times (0.34)^2 \][/tex]

[tex]\[ I_2 = \frac{2}{5}m_2r^2 = \frac{2}{5} \times 0.23 \times (0.34)^2 \][/tex]

Now, calculate the total moment of inertia:

[tex]\[ I_{\text{total}} = I_1 + I_2 \][/tex]

Substitute the values into the conservation of angular momentum equation:

[tex]\[ I_{\text{total}} \times ω_{\text{initial}} = I_{\text{total}} \times ω_{\text{final}} \][/tex]

Solve for [tex]\(ω_{\text{final}}\)[/tex] and then use it to calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball:

[tex]\[ v = ω_{\text{final}} \times r \][/tex]

After performing these calculations, we can determine the speed of the 100 g ball. Let me provide you with the numerical results in the next response.

After performing the calculations, the final angular velocity [tex](\(ω_{\text{final}}\))[/tex] is determined to be the same as the initial angular velocity [tex](\(4π \, \text{rad/s}\))[/tex]. Now, we can calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball using the formula [tex]\(v = ω_{\text{final}} \times r\).[/tex]

[tex]\[ v = 4π \times 0.34 \][/tex]

[tex]\[ v \approx 4 \times 3.14 \times 0.34 \][/tex]

[tex]\[ v \approx 4.24 \, \text{m/s} \][/tex]

Therefore, the speed of the 100 g ball in this rotating system is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]

Answer:

The sentence is following with "are made from mostly silicon material".

Explanation:

Silicon is extracted and refined by the use of electrical energy. The amount of energy required to produce the silicon in a solar cell is greater than the energy that cell will produce during its useful working life.

Answer:

262 kN/C

Explanation:

If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:

v² = v0² + 2aΔS

Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:

(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013

16x10¹⁴ = 4x10¹⁴ + 0.026a

0.026a = 12x10¹⁴

a = 4.61x10¹⁶ m/s²

The electric force due to the electric field (E) is:

F = Eq

Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:

F = m*a

Where m is the mass, so:

E*q = m*a

The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:

E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶

E = 261,866.42 N/C

E = 262 kN/C

Answer:

Resistance will become 4 times the previous value

Explanation:

We have given current in the circuit i = 2.4 A

According to ohm's law current in the circuit is given by [tex]I=\frac{V}{R}[/tex]

So [tex]2.4=\frac{V}{R}[/tex]............eqn 1

Now voltage is increased to 4 times so new voltage = 4 V

And current in the circuit is same as 2.4 A

We have to fond the resistance so that after increasing voltage current will be same

So [tex]2.4=\frac{4V}{R_{unknown}}[/tex]..........eqn 2

Dividing eqn 1 and 2

[tex]1=\frac{V}{R}\times \frac{R_{unkown}}{4V}[/tex]

[tex]R_{unknown}=4R[/tex]

So resistance will become 4 times the previous value

Answer:

It should increase to four times its value.

Explanation:

Answer: a) 42Nm b) 8.4m/s

Explanation:

Impulse is defined as object change in momentum.

Since Force = mass × acceleration

F = ma

Acceleration is the rate of change in velocity.

F = m(v-u)/t

Cross multiply

Ft = m(v-u)

Since impulse = Ft

and Ft = m(v-u)... (1)

The object change in velocity (v-u) = Ft/m from eqn 1

Going to the question;

a) Impulse = Force (F) × time(t)

Given force = 14N and time = 3seconds

Impulse = 14×3

Impulse = 42Nm

b) The object change in velocity (v-u) = Ft/m where mass = 5kg

v-u = 14×3/5

Change in velocity = 42/5 = 8.4m/s

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

The change in entropy of the gas is 374.39 J/K.

The given parameters:

The heat of fusion of the ice is calculated as follows;

[tex]\Delta H = mL\\\\\Delta H = 0.3071 \times 333,000\\\\\Delta H = 102,264.3 \ J[/tex]

The change in entropy of the gas is calculated as follows;

[tex]\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{102,264.3}{273.15} \\\\\Delta S = 374.39 \ J/K[/tex]

Thus, the change in entropy of the gas is 374.39 J/K.

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Answer:

increase

Explanation:

Electric potential is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. Taking a test positive charge from one point to another means that work is done against the field hence electric potential increases.

Answer: Jane 6ft from fulcrum

John 4 ft from fulcrum

Explanation: You use the law of moment of force.

That is Clockwise moment equals anti clockwise moment.

Here is the attachment involving the diagram and procedure of calculations and the answers.

The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.

The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.

Answer:

I think the answer is c

Explanation:

But it is kinda of opinionated

Statements C best describe a leader. A leader is a person who inspires other team members to reach the organizational goals

A leader is someone who motivates others to achieve the organization's objectives.

A leader is the person who owns the company and manages it on a daily basis. Any member of the management team might be considered a leader, but only a leader would be able to own the company.

A leader is a person who inspires other team members to reach the organizational goals

Hence,statements C best describe a leader.

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Explanation:

Formula for final volume of chamber if the partition is ruptured will be as follows.

        [tex]V_{2}[/tex] = 1.5 + 1.5

                   = 3.0 [tex]ft^{3}[/tex]

As mass remains constant then the specific volume at this state will be as follows.

             [tex]\nu_{2} = \frac{V_{2}}{m}[/tex]

                          = [tex]\frac{3.0}{2}[/tex]

                          = 1.5 [tex]ft^{3}/lbm[/tex]

Now, at final temperature [tex]T_{2}[/tex] = 300 F according to saturated water tables.

   [tex]\nu_{f} = 0.01745 ft^{3}/lbm[/tex]

   [tex]\nu_{fg} = 6.4537 ft^{3}/lbm[/tex]  

   [tex]\nu_{g} = 6.47115 ft^{3}/lbm[/tex]

Hence, we obtained [tex]\nu_{f} < \nu_{2} < \nu_{g}[/tex] and the state is in wet condition.

       [tex]\nu_{2} = \nu_{f} + x_{2}\nu_{fg}[/tex]

             1.5 = [tex]0.01745 + x_{2} \times 6.4537[/tex]

        [tex]x_{2}[/tex] = 0.229

Now, the final pressure will be the saturation pressure at [tex]T_{2}[/tex] = 300 F

and,   [tex]P_{2}[/tex] = [tex]P_{sat}[/tex] = 66.985 psia

Formula to calculate internal energy at the final state is as follows.

         [tex]U_{2} = m(u_{f}_{300 F} + x_{2}u_{fg_{300 F}}[/tex]

                   = [tex]2(269.51 + 0.229 \times 830.45)[/tex]

                   = 920.56 Btu

Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.

Answer:

Tribe historians would: Depict a significant event for each year on a buffalo or deer skin

Tribes of the Sioux Nation maintained historical calendars composed of winter counts, using pictorial representations to record significant events. Winter counts served as a visual record of the tribe's history and were passed down through generations.

Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would use pictorial representations to record significant events that occurred during each year. Each year, a new pictorial symbol would be added to the count. For example, if a significant battle occurred during a year, the historian would draw a symbol representing that battle. Winter counts served as a visual record of the tribe's history and were often passed down from generation to generation.

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Traction loss in a vehicle's rear wheels is most likely due to either acceleration-induced traction loss or rear wheel traction loss (skid). The former is caused by rapid acceleration, while the latter can be due to turning or braking.

The traction loss described in your question occurs in the rear wheels of a vehicle and is likely due to acceleration-induced traction loss or rear wheel traction loss (skid). Acceleration-induced traction loss happens when rapid acceleration causes the tires to lose grip on the road. This is common in high-powered, rear-wheel drive vehicles. Rear wheel traction loss or a rear wheel skid, on the other hand, generally occurs when the rear wheels lose grip during turning or braking. In this scenario, the vehicle's rear end can swing out in either direction, causing the vehicle to spin.

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Final answer:

Rear wheel traction loss (skid) occurs when the rear tires lose grip on the road surface. In a front-wheel-driven car, friction at the rear wheel is in the opposite direction of motion. This concept is important for analysis in vehicle dynamics and skid prevention.

Explanation:

The traction loss referred to in the question occurs in the rear wheels of a vehicle and is specifically known as rear wheel traction loss (skid). This is a type of skid that happens when the rear tires of a vehicle lose grip on the road surface, often as a result of oversteering, sudden acceleration, or slippery conditions. It's essential to distinguish this from other types of traction loss, such as braking-induced traction loss, which occurs when the tires can no longer grip the road during heavy braking, or driver-induced skids, which are the result of driver error.

With respect to the problem provided, in a front-wheel-driven car, the friction at the rear wheel is generally in the opposite direction of motion. This is because the rear wheels are not powered in this configuration; instead, they follow the rotation induced by the car's movement, which is primarily driven by the front wheels. Therefore, while the front wheels are pulled forward by the engine, creating friction in the direction of motion, the rear wheels experience a kind of rolling resistance and the friction present there acts in the opposite direction of the car's movement.

Understanding the dynamics of friction and traction is critical for analyzing cases like where a vehicle skids and comes to rest after traveling a certain distance or determining the force required to maintain a constant speed in the presence of kinetic friction. These analyses involve concepts like travel reduction ratio, torque transferred to the wheel axle, tractive efficiency of the wheel, and the actual velocity of the wheel.

Answer:

[tex]u_x=8.5454\ m.s^{-1}[/tex]

[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.

Explanation:

Given:

[tex]r=3.2\times 5[/tex]

[tex]r=16\ m[/tex]

Since the projectile was thrown horizontally therefore the vertical component of velocity is zero.

Now the time taken for the object to reach to Henrietta ignoring the height where she catches:

[tex]h=u_x.t+\frac{1}{2} \times g.t^2[/tex]

[tex]43.9=0+\frac{1}{2} \times 9.8\times t^2[/tex]

[tex]t=2.9932\ s[/tex] is the time taken by the projectile to reach Henrietta.

Now the distance further walked by Henrietta in the above time from the point where she was when the projectile was launched:

[tex]\Delta d=v_w\times t[/tex]

[tex]\Delta d=9.5782\ m[/tex]

Now the total  horizontal distance from the window to Henrietta when the projectile reached her:

[tex]d=\Delta d+r[/tex]

[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.

Now the initial horizontal velocity of launch of the projectile:

[tex]u_x=\frac{d}{t}[/tex]

[tex]u_x=\frac{25.5782}{2.9932}[/tex]

[tex]u_x=8.5454\ m.s^{-1}[/tex]

Explanation:

The life span of a star is quite as many more than 10 billion years. During these many years it would pass through different stages of its life,  from nebula to ( perhaps, the current stage) supernova to neutron star or Black Hole.

The stages of the life of star over a span of 13 billion year can be summarized as

the transition stages are

Nebula → Blue star → Blue-white super-giant  star → Supernova → Neutron star →Black Hole.

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

Answer:

We say that the object's spectrum is shifted.

Explanation:

Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

That shift can be used to find the velocity of the object by means of the Doppler velocity.

[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex]  (1)

Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.

The object's spectrum is affected by the Doppler effect due to its motion away from us.

The apparent change in the wavelength of a spectral line from a distant object compared to the laboratory measurement is known as the Doppler effect. In this case, the spectral line appears at a longer wavelength (328 nm) in the spectrum of the distant object compared to the laboratory measurement (321 nm), indicating that the object is moving away from us.

The Doppler effect occurs because when an object moves away from an observer, the wavelength of the light it emits appears to increase, causing a shift toward the red end of the spectrum. On the other hand, if the object was moving toward us, the spectral line would appear at a shorter wavelength and shift toward the blue end of the spectrum.

This phenomenon is important in astronomy as it allows us to determine whether distant objects like stars are moving toward or away from us, which helps us understand their motion and the expansion of the universe.

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Answer:

a.  [tex]W_w=313.5\ J[/tex]

b. [tex]W_g=-155.312\ J[/tex]

c. [tex]F_N=222.045\ N[/tex]

d.  [tex]W_t=313.5\ J[/tex]

Explanation:

Given:

a.

Work done by the worker who applied the force:

[tex]W_w=F.s\ cos 0^{\circ}[/tex] since the direction of force and the displacement are the same.

[tex]W_w=209\times 1.5[/tex]

[tex]W_w=313.5\ J[/tex]

b.

Work done by the gravitational force:

[tex]W_g=m.g\times h[/tex]

where:

g = acceleration due to gravity

h = the vertically downward displacement

Now, we find the height:

[tex]h=s\times sin\ \theta[/tex]

[tex]h=1.5\times sin\ 25^{\circ}[/tex]

[tex]h=0.634\ m[/tex]

So, the work done by the gravity:

[tex]W_g=25\times 9.8\times (-0.634)[/tex]  ∵direction of force and displacement are opposite.

[tex]W_g=-155.312\ J[/tex]

c.

The normal reaction force on the crate by the inclined surface:

[tex]F_N=m.g.cos\ \theta[/tex]

[tex]F_N=25\times 9.8\times cos\ 25[/tex]

[tex]F_N=222.045\ N[/tex]

d.

Total work done on crate is with respect to the worker: [tex]W_t=313.5\ J[/tex]

Answer:

The Tension T is 42120N

The Horizontal force component is 18322.2N

The Vertical force component is - 4729N

Explanation:

First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).

Having done that, you apply two conditions of equilibrium.

1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.

∑Fx = 0

∑Fx = Rx - Tcos64.2 = 0

Rx = 0.435T

∑Fy = 0

∑Fy = Ry + Tsin64.2 - W - w = 0

W = 2000kg × 9.8 = 19600N

w =1000kg × 9.8 = 9800N

Ry + 0.9T = 29400N

Ry = 29400 - 0.9T

2. THE SUM TOTAL OF TORQUES EQUALS ZERO

Rx: τ = 0

Ry: τ = 0

T: τ = 5 × Tsin44.2

= 3.49T m

W: τ = 4 × 19600sin90

= 78400Nm

w: τ = 7 × 9800sin9

= 68600Nm

Note:

Rx = x component of Reaction force

Ry = y component of Reaction force.

T = Tension

W = weight of bridge

w = weight of Sir Lance a Lost and his steed

τ = torque

Note: The torque of Tension is counter clockwise while that of the weights is clockwise.

Hence,

∑τccw = ∑τcw

3.49T = 78400 + 68600

3.49T = 14700Nm

T = 147000/3.49

T = 42120N

Rx = 0.435 × 42120

Rx = 18322.2N

Ry = 29400N - (0.9×42120)N

Ry = 29400 - 34129

Ry = -4729N

Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.

In this exercise we have to use the knowledge of tension and force, so we can say that this will result in:

A)The Tension T is 42120N

B)The Horizontal force component is 18322.2N

C)The Vertical force component is - 4729N

Before starting the calculations we have to remember some concepts such as:

First we will add all the force vectors so that it results in zero and that means that the body is in equilibrium, so:

[tex]\sum F_x = 0\\ \sum F_x = R_x - Tcos(64.2) = 0\\ R_x = 0.435T\\ \sum F_y = 0\\ \sum F_y = R_y + Tsin(64.2) - W - w = 0\\ W = 2000kg * 9.8 = 19600N\\ w =1000kg * 9.8 = 9800N\\ R_y + 0.9T = 29400N\\ R_y = 29400 - 0.9T[/tex]

Secondly, we will add up all the torques so that it results in zero and that means that the body is in equilibrium, so:

[tex]T: T = 5 * Tsin(44.2) = 3.49T m\\ W: T = 4 * 19600sin(90)= 78400Nm\\ w: T = 7 * 9800sin(90) = 68600Nm\\ \sum Tccw = \sum Tcw\\ 3.49T = 78400 + 68600\\ 3.49T = 14700Nm\\ T = 147000/3.49\\ T = 42120N R_x = 0.435 * 42120\\ R_x = 18322.2N\\ R_y = 29400N - (0.9*42120)N\\ R_y = 29400 - 34129\\ R_y = -4729N[/tex]

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Answer:

c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic

Explanation:

The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.

When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).

Answer: 0.006in/s

Explanation:

Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s

Also let the rate at which the radius of the balloon is increasing be dr/dt

Given r = 4.7in and Π = 3.14

Applying the chain rule method

dV/dt = dV/dr × dr/dt

If the volume of the sphere is 4/3Πr³

V = 4/3Πr³

dV/dr = 4Πr²

If r = 4.7in

dV/dr = 4Π(4.7)²

dV/dr = 277.45in²

Therefore;

1.68 = 277.45 × dr/dt

dr/dt = 1.68/277.45

dr/dt = 0.006in/s

Answer:

Range, [tex]R = MV²/2QE[/tex]

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

Answer:

a) 200m

b) 40 m/s

c) 81.55m

d) 31.62m

Explanation:

Solution

a)

y = y0 + u×t+⅟2×a×t2 =

y0 = 0

u = 0

y = unknown

a = 4m/s2

t = time = 10 seconds

y = 0.5×4×102 = 200m

b) v = u + at

v = 0 + 4×10 = 40 m/s

c) v2 = u2 - 2×g×y

at maximum height  v = 0

we have

402 = 2×9.81×y

Y =81.55m

d)

for the stage 2 we haace

 y = y0 + u×t+⅟2×a×t2 =

y = 0 + 4×2+0.5×6×22 = 92m

for the stage one we have

y = 0+40×2-0.5×9.81×4= 60.38m

distance between the first and second stage  2s aftee separation = 92-60.38 = 31.62m

Answer:

increase

Explanation:

think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing

Answer:

decrease.

Explanation:

The air pressure is given by

P = h x d x g

where, h is the height of air column above the surface and g be the acceleration due to gravity and d be the density of air.

As we go up, the height of air column decreases, density of air decreases and acceleration due to gravity also decreases, so the value of pressure decreases at altitudes.

Answer:

Velocity of rocket will be equal to 7.81 m/sec

Explanation:

We have given mass of the rocket [tex]m_1=4kg[/tex]

Mass of fuel is given [tex]m_2=50gram=0.05kg[/tex] ( As 1 kg is equal to 1000 gram )

Velocity of fuel [tex]v_2[/tex] = 625 m/sec

We have to find the velocity of rocket [tex]v_1[/tex]

From conservation of momentum we know that initial momentum is equal to final momentum '

So [tex]m_1v_1=m_2v_2[/tex], here [tex]m_1[/tex] is mass of rocket [tex]v_1[/tex] is velocity of rocket [tex]m_2[/tex] is mass of fuel and [tex]v_2[/tex] is velocity of fuel

So [tex]4\times v_1=625\times 0.05[/tex]

[tex]v_1=7.81m/sec[/tex]

So velocity of rocket will be equal to 7.81 m/sec

Answer:

20.6m

Explanation:

The mass is useless in this case.  

Kinetic Energy = Potential Energy

g is acceleration due to gravity.

(1/2)mv² = mgh (m's drop out)

(1/2)v² = gh

(1/2)(20.1²) = 9.81(h)

h = (20.1 x 20.1)/9.81 x 2

20.61 meters

Answer:

b. fluid friction occurs

Explanation:

Water is a liquid and both going through air gas and liquid cause fluid friction.