pls someone should help me answer question 3. Thank you​

Answer:

(a) 12 m/s

(b) At t = 0, x'(t) = 0

At t = 5.0 s, x'(t) = 15 m/s

At t = 10.0 s, x'(t) = 12 m/s

(c) i. The initial velocity = 2.0 cm/s

The initial position = 50 cm

The initial acceleration = 0.125 cm/s²

ii. 16 seconds

2. (a) Approximately 0.85 m/s

(b) 2.094 J

(c) i. Approximately 49.74 m/s

ii. Approximately 86.38°

3. (i) 4·i + 5·j

(ii) -2·i - j - 2·k

(iii) 5·i + 4·j - 3·z

(iv) 8

(v) (a) The magnitude is 2.8 cm, the direction is East

(b) The magnitude is ((14·√3)/5) cm, the direction is North

(c)  The magnitude is ((14·√3)/5), the direction is South

Explanation:

x(t) is given as follows;

x(t) = b·t² - c·t³

Where;

b = 2.4 m/s(²) (we note that the unit of b for the term to be distance in m/s²)

c = 0.120 m/s³

(a) At t = 0, we have;

x(0) = b × 0² - c × 0³ = 0

At t = 10.0 s, we have;

x(10.0) = 2.4 m/s² × (10.0 s)² - 0.120 m/s³ × (10.0 s)³ = 120 m

The average velocity = (Total distance)/(Total time)

∴ The average velocity of the car for the time interval t = 0 to t = 10.0 s, [tex]v_{ave}[/tex], is given as follows;

[tex]v_{ave}[/tex] = (120 m - 0 m)/(10.0 s - 0 s) = 12 m/s

The average velocity of the car for the time interval t = 0 to t = 10.0 s, [tex]v_{ave}[/tex]  = 12 m/s

(b) The instantaneous velocity, [tex]v_{inst}[/tex], is given as follows;

[tex]v_{inst} = \lim \limits_{t \to 0} \left( v_{ave}\right) = \lim \limits_{t \to 0} \left( \dfrac{\Delta x}{\Delta t} \right) = \dfrac{dx}{dt} = x'(t)[/tex]

[tex]x'(t) = \dfrac{d\left (b \cdot t^2 - c \cdot t^3\right)}{dt}[/tex]

x'(t) = 2·b·t - 3·c·t²

At t = 0, x'(t) = 2 × b × 0 - 3 × c × 0² = 0

At t = 5.0 s, x'(t) = 2 × 2.4 m/s² × 5.0 s - 3 × 0.120 m/s³ × (5.0 s)² = 15 m/s

At t = 10.0 s, x'(t) = 2 × 2.4 m/s² × 10.0 s - 3 × 0.120 m/s³ × (10.0 s)² = 12 m/s

(c) x(t) = 50 cm + (2.0 cm/s)·t - (0.0625 cm/s²)·t²

i. The initial velocity is the instantaneous velocity, x'(t), at time, t = 0

x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t

At t = 0, x'(0) = 2.0 cm/s - 2 × 0.0625 cm/s² × 0 = 2.0 cm/s

The initial velocity, x'(0) = 2.0 cm/s

The initial position = The position at time t = 0 = x(0)

x(0) = 50 cm + (2.0 cm/s) × 0 - (0.0625 cm/s²) × 0² = 50 cm

The initial position, x(0) = 50 cm

The initial acceleration, x''(0) = 2 × 0.0625 cm/s² = 0.125 cm/s²

ii. x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t

When the velocity of the turtle, x'(t) = 0 we have;

0 = 2.0 cm/s - 2 × 0.0625 cm/s² × t

∴ t = (2.0 cm/s)/(2 × 0.0625 cm/s²) = 16 seconds

The velocity of the turtle is zero after 16 seconds

2. The mass of the large fish, m₁ = 15.0-kg

The speed of the large fish, v₁ = 1.1 m/s

The mass of the smaller fish, m₂ = 4.50 kg

The speed of the small (stationary) fish, v₂ = 0

The initial momentum = 15.0 kg × 1.1 m/s + 4.50 kg × 0 = 16.5 kg·m/s

The initial momentum = 16.5 kg·m/s

The final momentum = (15.0 kg + 4.50 kg) × v₃ = 19.50 kg × v₃

The final momentum = 19.50 kg × v₃

Where;

The total initial momentum = The total final momentum

We get;

16.5 kg·m/s = 19.50 kg × v₃

∴ v₃ = (16.5 kg·m/s)/(19.50 kg)

v₃ = (16.5/19.50) m/s = (11/13) m/s ≈ 0.85 m/s

∴ The speed of the large fish just after it eats the small, v₃ ≈ 0.85 m/s

(b) The initial kinetic energy, K.E.₁ = (1/2) × 15 kg × (1.1 m/s)² = 9.075 J

The final kinetic energy, K.E.₂ = (1/2) × 19.5 kg × (11/13 m/s)² = 363/52 J

The mechanical energy dissipated, ΔE = K.E.₁ - K.E.₂

ΔE = (9.075 - 363/42) J = 1089/520 J ≈ 2.094 J

The mechanical energy dissipated, ΔE = 2.094 J

(c) i. We have the total momentum = 110 × 8.8· j + 85 × 7.2· i = 9.680·i + 612·i

The velocity after collision, v = (9.680·i + 612·i)/(110 + 85) = 49.64·j + 3.14·i

The magnitude of the velocity, v = √(49.64² + 3.14²) ≈ 49.74 m/s

ii. The direction, θ = arctan(49.64/3.14) ≈ 86.38°

3. (i) [tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = (i + 2·j - k) + 3·i + 3·j + k = 4·i + 5·j

(ii) [tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = (i + 2·j - k) - (3·i + 3·j + k) = -2·i - j - 2·k

(iii) [tex]\underset{A}{\rightarrow} \times \underset{B}{\rightarrow}[/tex] = (2 + 3)·i - (1 + 3)·j + (3 - 6)·z = 5·i + 4·j - 3·z

(iv) [tex]\underset{A}{\rightarrow} \cdot \underset{B}{\rightarrow}[/tex] = 1×3 + 2 × 3 + (-1)×1 = 8

(v) (a) [tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = 2.8·cos(60°)·i + 2.8·sin(60°)·j + (2.8·cos(60°)·i - 2.8·sin(60°)·j

[tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = 5.6·cos 60°·i = 2.8·i

The magnitude = 2.8, the direction is east

(b) [tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = 2.8·cos(60°)·i + 2.8·sin(60°)·j - (2.8·cos(60°)·i - 2.8·sin(60°)·j

[tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = 5.6·sin(60°)·j= ((14·√3)/5)·j

The magnitude = ((14·√3)/5), the direction is North

(c)  [tex]\underset{B}{\rightarrow} - \underset{A}{\rightarrow}[/tex] = (2.8·cos(60°)·i - 2.8·sin(60°)·j - (2.8·cos(60°)·i + 2.8·sin(60°)·j)

[tex]\underset{B}{\rightarrow} - \underset{A}{\rightarrow}[/tex] = -5.6·sin(60°)·j= (-(14·√3)/5)·j

The magnitude = ((14·√3)/5), the direction is South

The waste product of photosynthesis is

Answer pls…………………………

What are various systems of unit?​

What is 3.75 x 10^-7?

what are the 3 formulas which describe the relationship between mass force and acceleration

i have 5 eggs i grabbed one i a cracked 1 i cooked one i ate one how many eggs do i have

Hey guys Im Anushka from India... pls help me out in this question..

What is linear motion?

What is rotational motion?​

When do we have positive/negative and zero acceleration.Write down the terms..................pls help mee

A 15 cm length of wire is moving perpendicularly
through a magnetic field of strength 1.4 T at the rate
of 0.12 m/s. What is the EMF induced in the wire?
A. OV
C. 0.025 v
B. 0.018 V
D. 2.5 V

Explain the energy transformations that occur when breaking in a hybrid
vehicle.

trong chân không có 2 điện tích điểm q1=3.10^-6C và q2=-6.10^-6C đặt tại A và B tương ứng cách nhau 100cm. Tính lực điện thành phần và tổng hợp do q1, q2 tác dụng lên q0=5.10^-6C đặt tại M

A small block is attached to a spring with a spring constant of 85 N/m. When the spring is compressed 0.30 meters and the released, the 0.50 kg block is launched. What is the final speed of the block?


A)
8.42 m/s

B)
3.91 m/s

C)
7.14 m/s

D)
15.3 m/s

The density of oil is 0.8 g/cm^3. What is the mass of 36 cm^3 of oil?​

Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum constant velocity C. Not in motion D. Point of action of weight E. Turning effect of force F. Study of forces 4. Terminal velocity 5. Statics​

the same crops are grown in different seasons true or false​

On which planet would your weight be the most and the least?

a.Jupiter and Mercury
B.Jupiter and Neptune
c.Saturn and Neptune
D.
Saturn and Uranus

in a series circuit, how does the voltage supplied by the battery compare to the voltages on each load?

A meter 0.0525 m from a wire measures a magnetic field of 7.14 x 10-6 T. How much current flows through the wire?​

The frequency of a wave is the inverse of the wave's
wavelength
itude
perod
velocit

Which of the following are
matter?
Chair, air, love, smell, hate,almonds, thought, cold, cold-drink, smell of perfume.​

I need help with question 6

Please answer in your own words and I will mark brainlist, Compare all matters from the view of passing light through them (name 3 types and write if they pass the light or not.)​

How to calculate displacement, velocity, acceleration.

state the factor that affect the gravity of a planet​

A car travelling along a straight road has kinetic energy of 180 kJ. The brakes are applied, and it is brought to rest over a distance of 60 m. What is the average force of the car breaks?
A. 3 N

B. 3000 N

C. 10,800 N

explain why heating destroys the magnetism in a permanent magnet?​

When playing a game of disc golf, each throw to your target is considered to be a O Point O Stroke O Hit O Toss​

In a bicycle dynamo,does 1. The permanent magnet surrounds a conducting coil 2. The conducting coil rotates when the rear wheel of the bicycle rotates. 3.The electricity is generated in the permanent magnet 4.When the rear wheel of the bicycle turns fast, the brightness of the light increases

A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g = 10 m/s^2 find the maximum height reach by the stone and what is the net displacement and distance covered by the stone.​

Parallel incident rays appear to bounce like they have all originated from the same point. What is this point called?

A. cross point

B. midpoint

C. bounce point

D. focal point