Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

[tex]I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4[/tex]

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

[tex]\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa[/tex]

Calculate the first moment of area below point C

[tex]Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm[/tex]

Now calculate shear stress at point C

[tex]=\frac{FQ}{It}[/tex]

[tex]=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa[/tex]

Calculate the principal stress at point C

[tex]\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa[/tex]

Calculate the maximum shear stress at piont C

[tex]\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2} \\\\=336.28KPa[/tex]

Answer: 10m

Explanation:

The magnitude of the vector would be 10

[tex]\sqrt{6^{2}+8^{2} } =10[/tex]

Answer:

Vf = 78.64 m/s

Explanation:

The rocket is travelling upward at a constant acceleration of 3.99 m/s² until it runs out of fuel. So, in order to calculate its velocity at the point, where it runs out of fuel, we can simply use 3rd equation of motion:

2as = Vf² - Vi²

where,

a = acceleration = 3.99 m/s²

s = distance or height covered by rocket till fuel runs out = 775 m

Vf = Final Velocity = ?

Vi = Initial velocity = 0 m/s   (Since, rocket starts from rest)

Therefore,

2(3.99 m/s²)(775 m) = Vf² - (0 m/s)²

Vf = √(6184.5 m²/s²)

Vf = 78.64 m/s

Answer:

Explanation:

Givens

vi = 20 m/s

vf = 28 m/s

t = 2 seconds

Formula

a = (vf - vi) / t

Solution

a = (28 - 20)/2

a = 8/2

a = 4 m/s^2

Complete Question

A small car and an SUV are at a stoplight. The car has amass equal to half that of the SUV, and the SUV's engine can produce a maximum force equal to twice that of the car. When the light turns green, both drivers floor it at the same time. Which vehicle pulls ahead of the other vehicle after a few seconds?

a) It is a tie.

b) The SUV

c) The car

Answer:

The correct option is  a

Explanation:

From the question we are told that

     The mass of the car is [tex]m_c[/tex]

     The force of the car is  F

       The mass of the SUV is  [tex]m_s = 2 m_c[/tex]

       The force of the SUV is [tex]F_s = 2 F[/tex]

Generally force  of the car is mathematically represented as

        [tex]F= m_ca_c[/tex]

[tex]a_c[/tex] is acceleration of the car

Generally force  of the car is mathematically represented as

       [tex]F_s = m_s * a_s[/tex]

[tex]a_s[/tex] is acceleration of the SUV

=>   [tex]2 F = 2 m_c a_s[/tex]

       [tex]F = m_c a_s[/tex]

=>    [tex]m_c a_s = m_ca_c[/tex]

So  [tex]a_s = a_c[/tex]

  This means that the acceleration of both the car and the SUV are the same

Answer:

Explanation:

1 ) Since it is a isochoric process , heat energy passed into gas

= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .

= 7.4 x 12.47 x ( 500 - 300 )

= 18455.6 J.

2 ) Since there is no change in volume , work done by the gas is constant.

3 ) from  , gas law equation

PV = nRT

P = nRT / V

= 7.4 x 8.3 x 500 / .74

= .415 x 10⁵ Pa.

4 ) Average kinetic energy  of gas molecules after attainment of final temperature

= 3/2 x R/ N x T

= 1.5 x 1.38  x 10⁻²³ x 500

= 1.035 x 10⁻²⁰ J

1/2 m v² = 1.035 x 10⁻²⁰

v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶

= 1.49 x 10⁶

v = 1.22 x 10³ m /s

5 )  In this process , pressure remains constant

gas is cooled from 500 to 300 K

heat will be withdrawn .

heat withdrawn

= n Cp dT

= 7.4 x 20.79 x 200

= 30769.2 J .

6 )

gas will have reduced volume due to cooling

reduced volume = .74 x 300 / 500

= .444 m³

change in volume

= .74 - .444

= .296 m³

work done on the gas

= P x dV

pressure x change in volume

= .415 x 10⁵ x .296

= 12284 J.

Answer:2000

Explanation:

Answer:

0

Explanation:

if the volume remains constans, the works is 0 because the equation

W = P . ∆V

P = pressure

∆V = change in volume

Answer: Short Answer: NO ( In Most Cases)

Explanation:

If that were true then planes couldn't get off the ground to fly. The front of the wing is cutting/pushing the air. On the top of the wing the air moves faster and on the bottom it moves slower making a upward draft giving the object the ability to fly or glide.

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)

W is described as: W = m.g

Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N

N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:

[tex]W_y[/tex] = m.g.cos(14)

Therefore, the formula will be:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

Answer:

The average velocity of Jackson is 18.056 m/s South

The average velocity of Hunter is 10.65 m/s East

Explanation:

initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east

time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds

⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m

Final velocity of Jackson, v =  45 km/h South = 12.5 m/s South

time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes

time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s

⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m

Average velocity of Jackson;

[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]

initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South

time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds

⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m

Final velocity of Hunter, v =  40 km/h east = 11.11 m/s east

time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes

time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s

⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m

Average velocity of Hunter;

[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]

Answer:

-2.24 m/s²

Explanation:

Given:

v₀ = 20 mi/hr = 8.94 m/s

v = 0 m/s

t = 4.0 s

Find: a

v = v₀ + at

0 m/s = 8.94 m/s + a (4.0 s)

a = -2.24 m/s²

Answer:weight

Explanation:weight

Answer: B, Companies passed on production and transportation costs to consumers

Explanation:

A higher oil price occurred when companies passed on production and transportation costs to consumers.

The oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.

The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.

Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.

Learn more about inflations here: https://brainly.com/question/1082634

Answer:

B. energy

Explanation:

A vector has direction.

Energy does not have a direction.

Answer:

6.4 m/s

Explanation:

From the question, we are given that

Speed of the river, v(r) = 5 m/s

velocity relative to the water, v(w) = 4 m/s

Width of the river, d = 780 m

The magnitude of his velocity relative to the earth is v(m)

v(m) can be gotten by using the relation

[v(m)]² = [v(w)]² + [v(r)]²

[v(m)]² = 4² + 5²

[v(m)]² = 16 + 25

[v(m)]² = 41

v(m) = √41

v(m) = 6.4 m/s

thus, the magnitude of the velocity relative to earth is 6.4 m/s

Answer:

The correct option is D: "The small car and the truck experience the same average force."

Explanation:

The magnitude of the average force experienced by both bodies in motion is the same as explained by Newton's third law of motion. The force exerted by each body is equal and opposite in direction. The resulting acceleration experienced by each vehicle, however, will not be the same. It is greater for the small car.

Answer:

Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

This answer will approach this question in two steps:

For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.

The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:

[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].

The idea is to solve for the final angular velocity using the angular analogy of that equation:

[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].

In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:

Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:

[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].

Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:

[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].

The size of this force should be equal to that of the centripetal force when the car is about to skid:

[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].

Solve this equation for [tex]\mu_s[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].

Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].

(Three significant figures.)

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

Answer:

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Explanation:

The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-

We represent

The ball mass =  m = 275 g = 0.275 kg

Thus it goes to the floor and resurfaces upward.

The ball hits the ground at 3.30 m/s speed that is

u = -3.30 m/s which represents the Negative since the ball hits the ground)

It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)

[tex]\Delta p=p_f-p_i[/tex]

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))[/tex]

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Answer:

The difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction, or "forward." In AC, electrons keep switching directions, sometimes going "forward" and then going "backward."

Answer:

32N

Explanation:

The Left force exerts an opposite horizontal force of 32N on the middle object

Answer:

The power is  [tex]P_p = 4.6 \ W[/tex]

Explanation:

From the question we are told that

   The power delivered is  [tex]P_{s} = 1.15 \ W[/tex]

   Let it resistance be denoted as R

    The resistors are connected in series so the equivalent resistance is  

     [tex]R_{eqv} = R+ R = 2 R[/tex]

Considering when it is connected in series    

Generally power is mathematically represented as

     [tex]P_s = V * I[/tex]

Here I is the current which is mathematically represented as

       [tex]I = \frac{V}{2R}[/tex]

The power becomes

     [tex]P_s = V * \frac{V}{2R}[/tex]

     [tex]P_s = \frac{V^2}{2R}[/tex]

substituting value

    [tex]1.15 = \frac{V^2}{2R}[/tex]

Considering when resistance is connected in parallel

  The equivalent resistance becomes

    [tex]R_{eqv} = \frac{R}{2}[/tex]

So The current  becomes

       [tex]I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}[/tex]

And the power becomes

     [tex]P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s[/tex]

 substituting values

     [tex]P_p = 4 * 1.15[/tex]

     [tex]P_p = 4.6 \ W[/tex]

Answer:

F = 0.314 N

Explanation:

In order to calculate the applied force to the ball by the ground, you first calculate the speed of the ball just before it hits the ground. You use the following formula:

[tex]v^2=v_o^2+2gy[/tex]        (1)

y: height from the ball starts its motion = 1.8 m

vo: initial velocity = 0 m/s

g: gravitational acceleration =  9.8 m/s^2

v: final velocity of the ball = ?

You replace the values of the parameters in the equation (1):

[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(1.8m)}=5.93\frac{m}{s}[/tex]

Next, you take into account that the force exerted by the ground on the ball is given by the change, on time, of the linear momentum of the ball, that is:

[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}=m\frac{v_2-v_1}{\Delta t}[/tex]      (2)

m: mass of the ball = 20g = 20*10^-3 kg

v1: velocity of the ball just before it hits the ground = 5.93m/s

v2: velocity of the ball after it impacts the ground (80% of v1):

0.8(5.93m/s) = 4.75 m/s

Δt: time interval o which the ground applies the force on the ball = 75*10^-3 s

You replace the values of the parameters in the equation (2):

[tex]F=(20*10^{-3}kg)\frac{4.75m/s-5.93m/s}{75*10^{-3}s}=-0.314N[/tex]

The minus sign means that the force is applied against the initial direction of the motion of the ball.

The applied force by the ground on the bouncy ball is 0.314 N

Answer:

To determine wavelength of a wave on a string.

Explanation:

The Standing Waves lab study the parameters that affect standing waves in various strings. The effects of string tension and density on wavelength and frequency will be studied.

Answer:

6 gallons

Explanation:

At 30 mph, the fuel mileage is 25 mpg.

After 5 hours, the distance traveled is:

30 mi/hr × 5 hr = 150 mi

The amount of gas used is:

150 mi × (1 gal / 25 mi) = 6 gal

Answer:

The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.

Explanation:

The formula for electric field is given as:

E = F/q

where,

E = Electric field

F = Electric Force

q = Charge Experiencing Force

Now, we apply this formula to all the cases given in question.

A) A 2C charge acted on by a 4 N electric force

F = 4 N

q = 2 C

Therefore,

E = 4 N/2 C = 2 N/C

B) A 3 C charge acted on by a 5 N electric force

F = 5 N

q = 3 C

Therefore,

E = 5 N/3 C = 1.67 N/C

C) A 4 C charge acted on by a 6 N electric force

F = 6 N

q = 4 C

Therefore,

E = 6 N/4 C = 1.5 N/C

D) A 2 C charge acted on by a 6 N electric force

F = 6 N

q = 2 C

Therefore,

E = 6 N/2 C = 3 N/C

E) A 3 C charge acted on by a 3 N electric force

F = 3 N

q = 3 C

Therefore,

E = 3 N/3 C = 1 N/C

F) A 4 C charge acted on by a 2 N electric force

F = 2 N

q = 4 C

Therefore,

E = 2 N/4 C = 0.5 N/C

The highest field is 3 N, which is found in part D.

A 2 C charge acted on by a 6 N electric force

Answer:

Explanation:

  a )

from lens makers formula

[tex]\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})[/tex]

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

[tex]\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})[/tex]

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}[/tex]

[tex]\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}[/tex]

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

[tex]T=2\pi\sqrt{\frac{m_c}{k}}[/tex]     (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

[tex]m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg[/tex]    (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

[tex]T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg[/tex] (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

[tex]m_a=M-m_c=60.79kg-12.31kg=48.48kg[/tex]

The mass of the astronaut is 48.48 kg

Answer:

a) 98 m/s

b) 19.6 s

c)  1449.8 m

d)  29.6 s

e)  168.6 m/s

f)  46.8 s

Explanation:

Given that

Acceleration of the rocket, a = 5 m/s²

Altitude of the rocket, s = 960 m

a)

Using the equation of motion

v² = u² + 2as, considering that the initial velocity, u is 0. Then

v² = 2as

v = √2as

v = √(2 * 5 * 960)

v = √9600

v = 98 m/s

b)

Using the equation of motion

S = ut + ½at², considering that initial velocity, u = 0. So that

S = ½at²

t² = 2s/a

t² = (2 * 960) / 5

t² = 1920 / 5

t² = 384

t = √384 = 19.6 s

c)

Using the equation of motion

v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that

0 = 98² + 2(-9.8) * s

9600 = 19.6s

s = 9600/19.6

s = 489.8 m

The maximum altitude now is

960 m + 489.8 m = 1449.8 m

d)

Using the equation of motion

v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that

0 = 98 +(-9.8 * t)

98 = 9.8t

t = 98/9.8

t = 10 s

Total time then is, 10 + 19.6 = 29.6 s

e) using the equation of motion

v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,

v² = 0 + 2 * 9.8 * 1449.8

v² = 28416.08

v = √28416.08

v = 168.6 m/s

f) using the equation of motion

S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s

1449.8 = 0 + ½ * 9.8 * t²

2899.6 = 9.8t²

t² = 2899.6/9.8

t² = 295.88

t = √295.88

t = 17.2 s

total time in air then is, 17.2 + 29.6 = 46.8 s