. A ball weighs 120g on the earth surface,

i) What is its mass on the surface of the moon? 1mk

​

Answer:

a

The free body diagram is shown on the first uploaded image

b

The tension on the rope is  [tex]T=39.16 \ N[/tex]  

Explanation:

From the  question we are told that

    The mass of the box is  m

    The tension on the box is  T

     The angle at which it is pulled is  [tex]\theta = 40^o[/tex]

     The force produced by the strong head wind is [tex]Fw = 30 \ N[/tex]

At equilibrium the net force acting on the block along the horizontal axis is zero i.e

     [tex]Tcos \theta -F_w = 0[/tex]

substituting values

     [tex]Tcos (40) -30 = 0[/tex]

     [tex]Tcos (40) = 30[/tex]

     [tex]T(0.76604)) = 30[/tex]

     [tex]T=39.16 \ N[/tex]      

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

Answer:

Explanation: please see attached file I attached the answer to your question.

The angle between the magnetic field and the wire's velocity is 33.2 degrees.

Since the potential difference = 71mv = 71 *10 ^-3 V

The length is 12 cm = 0.12m

The magnetic field i.e. B = 0.27T

The speed or v = 4 m/s

here we assume [tex]\theta[/tex] be the angle

So,

e = Bvl sin[tex]\theta[/tex]

So,

[tex]Sin\theta[/tex] = e/bvl

= 71*10^-3 / 0.27 *4*0.12

= 0.5478

= 33.2 degrees

Therefore, the angle should be 33.2 degrees

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Answer:

The correct answer is a

Explanation:

The cosine function is

      cos θ = ca / ​​H

done ca is the adjacent leg (x-axis) and H is the hypotenuse (vector module)

we clear

    H = ca / ​​cos θ

therefore, to find the magnitude of the vector, the cathete is divided into the cosine.

The correct answer is a

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.

(A) The wheel has angular velocity at time [tex]t[/tex] according to

[tex]\omega=\omega_0+\alpha t[/tex]

so that after 2.50 s, the wheel will have attained an angular velocity of

[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]

(B) The angular displacement of the wheel is given by

[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]

After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to

[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]

Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

Answer:

The initial angular speed is [tex]w_i = 48 \ rad/s[/tex]

Explanation:

From the question we are told that

   The angular displacement is  [tex]\theta = 220 \ rad[/tex]

    The angular speed is  [tex]w_f = 92 \ rad/s[/tex]

    The acceleration is  [tex]\alpha = 14 \ rad/s^2[/tex]

Generally the initial angular speed can be evaluated as

     [tex]w_f ^2 = w_i ^2 + 2 * \alpha * \theta[/tex]

=>  [tex]w_i ^2 = w_f ^2 - 2 * \alpha * \theta[/tex]

substituting values

=>     [tex]w_i ^2 = 92 ^2 - 2 * 14 * 220[/tex]

=>      [tex]w_i ^2 = 2304[/tex]

=>     [tex]w_i = 48 \ rad/s[/tex]

Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒  Magnetic force = Copper wire's weight

So,

⇒   [tex]B\times I\times L=M\times g[/tex]

On putting the estimated values, we get

⇒  [tex]B\times 50\times 1=7.037\times 10^{-3}\times 9.81[/tex]

⇒  [tex]50B=69.03297\times 10^{-3}[/tex]

⇒  [tex]B=1.38\times 10^{-3} \ T[/tex]

(b)...

As we know,

⇒  [tex]m=\delta\times L\times \frac{\pi \ d^2}{4}[/tex]

⇒      [tex]=8960\times 1\times \frac{\pi \ (0.001)^2}{4}[/tex]

⇒      [tex]=2240\times \pi \ 0.000001[/tex]

⇒      [tex]=7.037\times 10^{-3} \ kg[/tex]

Answer:

Work-done in quarter an hour = 3.5 × 10⁶ J

Explanation:

Given:

Force (F) = 280 KN = 280,000 N

Velocity (V) = 50 km / h

Time (t) = 1 / 4 = 0.25 hour

Find:

Work-done in quarter an hour

Computation:

⇒ Displacement = Velocity (V) × Time

⇒ Displacement = 50 × 0.25

⇒ Displacement = 12.5 km

⇒ Work-done = Force (F) × Displacement

⇒ Work-done in quarter an hour = 280,000 × 12.5

⇒ Work-done in quarter an hour = 3,500,000

Work-done in quarter an hour = 3.5 × 10⁶ J

Answer:

d

Explanation:

good question. now the bus is moving in constant velocity . a student in front tosses a ball to the student in back. but we dont know the speed at which the student tosses a ball. we have to assume the speed

assume the speed of ball is slightly less than the speed of bus. in this case the stationary observer sees the ball in slower speed than the one inside the bus.

so a is correct

now assume the speed of ball is 1/2 the speed of bus. here stationary observer sees the ball the same speed as the one in bus observe

b is correct

assume the speed of ball is very small than the speed of bus . in this case the stationary observer see in grater speed than the student in bus

e also correct

so correct answer is d. it depends on the speed of ball tossed by the student in front.

Answer:

61.6° west of South

Explanation:

The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.

Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side  and the hypotenuse is 8

cos(θ) = [adjacent/hypotenuse]

Cos θ = 3.8/8

Cos θ = 0.475

θ = cos^-1 (0.475)

θ = 61.6°

Therefore the angle is 61.6° west of South.

Answer:

Final Temperature = 71 °C

Explanation:

In this case, the ideal gas equation is written as;

PV = mRT

Where;

P is pressure

V is volume

m is mass

R is gas constant

T is temperature

We will take the volume to be constant.

So, in the initial state, we have;

P1•V = m1•R•T1 - - - eq(1)

In the final state, we have;

P2•V = m2•R•T2 - - - - eq(2)

Combining eq (1) and eq(2),we have;

P1•m2•R•T2 = P2•m1•R•T1

Dividing both sides by R gives;

P1•m2•T2 = P2•m1•T1

Making T2 the subject gives;

T2 = (P2•m1•T1)/(P1•m2)

Now, we are given;

m1 = 2kg

m2 = ½*2 = 1kg

P1 = 4 atm

P2 = 2.2 atm

T1 = 40°C = 273 + 40 K = 313K

Plugging in this values into the T2 equation, we have;

T2 = (2.2 × 2 × 313)/(4 × 1)

T2 = 344 K

Converting to °C, we have;

T2 = 344 - 273 = 71 °C

Answer:-

-1 m/s^2

Explanation:

The average acceleration is given by dividing the change in velocity by change in time;

[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]

[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]

the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.

Answer:

A) E = 278925.62 N/C with direction; radially out.

B) E = 43048.47 N/C with direction radially out.

C) E = -3214.29 N/C with direction radially in.

Explanation:

From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;

E = kQ/r²

where;

Q is the net charge within the distance r.

We are given the charge Q = 15-nC and

spherical shell of radius 10cm

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C

B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C

C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Find out more information about magnitude here:

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Explanation:

We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of the pendulum

g is acceleration due to gravity on which it is placed

It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.

Answer:

The force constant is  [tex]k =1.316 *10^{7} \ N/m[/tex]

The energy stored in the spring is  [tex]E = 1.68 *10^{7} \ J[/tex]

Explanation:

From the question we are told that

   The mass of the object is  [tex]M = 4.8*10^{5} \ kg[/tex]

    The period is [tex]T = 1.2 \ s[/tex]

The period of the spring oscillation is  mathematically represented as

         [tex]T =2 \pi \sqrt{ \frac{M}{k}}[/tex]

where  k is the force constant

   So making k the subject

       [tex]k = \frac{4 \pi ^2 M }{T^2}[/tex]

substituting values

       [tex]k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}[/tex]

      [tex]k =1.316 *10^{7} \ N/m[/tex]

The energy stored in the spring is mathematically represented  as

       [tex]E = \frac{1}{2} k x^2[/tex]

Where x is the spring displacement which is given as

        [tex]x = 1.6 \ m[/tex]

substituting values

      [tex]E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2[/tex]

       [tex]E = 1.68 *10^{7} \ J[/tex]

Answer:

Explanation:

Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...

Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...

Answer:

a) R = 2.5 Ω, b) R = 1 Ω, c)     R = 2R / 3 Ω

Explanation:

The resistance configuration can be in series or in parallel, for each one the equivalent resistance can be calculated

series, the equivalent resistance is the sum of the resistances

parallel, the inverse of the equivalent resistance is the inverse of the sum of the resistances

let's apply these principles to each case

case a)

    equivalent series resistance

         R₁ = 1 +4 = 5 ohm

         R₂ = 2 +3 = 5 ohn

these two are in parallel

        1 / R = 1/5 +1/5

        1 / R = 2/5

         R = 2.5 Ω

case B

we solve the parallel

       1 / R₁ = ½ + ½ = 1

        R₁ = 1 Ω

we solve the resistors in series

       R₂ = 1 + 1

       R₂ = 2 Ω

finally we solve the last parallel

       1 / R = ½ +1/2 = 1

        R = 1 Ω

case C

we solve house resistance pair in series

     R₁ = R + 2R = 3R

we go to the next mesh

     R₂ = R + 2R = 3R

     R₃ = R + 2R = 3R

last mesh

     R₄ = R + R = 2R

now we solve the parallel of this equivalent resistance

     1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄

     1 / R = 1 / 3R + 1 / 3R + 1 / 3R + 1 / 2R

      1 / R = 3 / 3R + 1 / 2R = 1 / R + 1 / 2R

     1 / R = 3 / 2R

      R = 2R / 3 Ω

Answer:

Its initial position was 471 m.

Explanation:

We have,

Final position of the object is 327 m

Displacement of the object is -144 m

It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,

[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]

So, its initial position was 471 m.

Answer:

Q = 3877 KJ

Explanation:

Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:

Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)

Q = Q₁ + Q₂   ----- equation (1)

Now, for Q₁:

Q₁ = m C ΔT

where,

m = Mass of Water = 1.5 kg

C = Specific Heat of Water = 4187 J/kg.°C

ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C

Therefore,

Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)

Q₁ = 490 x 10³ J =490 KJ

Now, for Q₂:

Q₂ = m H

where,

m = Mass of Water = 1.5 kg

H = Heat of Vaporization of Water = 2258 KJ/kg

Therefore,

Q₂ = (1.5 kg)(2258 KJ/kg)

Q₂ = 3387 KJ

Substituting the values in equation (1), we get:

Q = Q₁ + Q₂

Q = 490 KJ + 3387 KJ

Q = 3877 KJ

Answer:

14,260

Explanation:

Relevant data provided for computing the wavelengths are in one pulse is here below:-

The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]

Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]

The number of wavelengths are in one pulse is shown below:-

[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]

[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]

= 14,260

Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.

Answer:

[tex]\large \boxed{42\, \mu \text{C}}$[/tex]

Explanation:

The formula for the force exerted between two charges is

[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

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

[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]

Answer:

[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

Explanation:

Angular acceleration

[tex]\begin{aligned}

\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\

\omega_{i} &=0 \\

\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\

&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\

&=27.02 \mathrm{rad} / \mathrm{s} \\

\alpha &=\frac{(27.02-0)}{3.15} \\

&=8.57 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

a)Tangential acceleration

[tex]\begin{aligned}

a &=r \alpha \\

&=\frac{12}{2} \times 10^{-2} \times 8.57 \\

a &=0.51 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

This question involves the concepts of the equations of motion for angular motion.

The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".

First, we will use the first equation of motion for the angular motion to find out the angular acceleration:

[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

where,

[tex]\alpha[/tex] = angular acceleration = ?

[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s

[tex]\omega_i[/tex] = initial angular speed = 0 rad/s

t = time taken = 3.05 s

Therefore,

[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]

Now, the tangential acceleration can be given as follows:

[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]

a = 0.532 m/s²

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The attached picture shows the angular equations of motion.

Answer:

K = 3.9 kJ

Explanation:

The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:

[tex] K_{T} = K_{f} - K_{i} [/tex]  

The initial kinetic energy is:

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]

Where m₁ is the mass of the object before the explosion and v₁ is its velocity

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]

Now, the final kinetic energy is:

[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]

Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces

Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex]   (1)          

By conservation of momentum we have:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]  

[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]

[tex] v_{2} = \frac{5}{4}v_{1} [/tex]     (2)

By entering (2) into (1) we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]  

Hence, the kinetic energy added is:

[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]  

Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.

I hope it helps you!

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

In the given case, the speed of the discus at release, If the thrower takes 1.2s to complete one revolution, starting from rest would be - 8.90 m/s.

Given:

radius f the circle would be = 1.7/2 = 0.85 m

This rotation exercise can be treated using the rotation kinematics.

Angular acceleration:

θ = w₀ t + ½ α t²

t = 1.2 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)

 =>  θ = 0 + ½ α t²

 => α = 2θ / t²

=>  α= 2 × 2π / 1.2²

 => α = 4π = 8.7266 rad / s²

Let's calculate the angular velocity:

=> w = wo + α t

=>  w = 0 + α t

=> w = 8.7266 × 1.2

=> w = 10.47192 rad / s

The relationship between linear and angular velocity is

=> r = d / 2

=> r = 1.7 / 2 = 0.85 m

=> v = w r

=> v = 10.47192 × 0.85  

=> v = 8.90 m / s

Thus, the correct speed would be - 8.90 m/s

Learn more:

https://brainly.com/question/14663644

Answer:

F = 423.63 N

Explanation:

Since, the x-component and y-components of the acceleration of ball are given. Therefore, we need to find the resultant or net acceleration of the soccer ball first. For that purpose we use to the formula for the resultant of rectangular components of a vector:

a = √(ax² + ay²)

where,

a = net acceleration = ?

ax = x - component of acceleration = 950 m/s²

ay = y - component of acceleration = 750 m/s²

Therefore,

a = √[(950 m/s²)² + (750 m/s²)²]

a = 1210.4 m/s²

Now, from Newton's Second Law, we know that:

F = ma

where,

m = mass of ball = 0.35 kg

F = Net force acting on ball = ?

F = (0.35 kg)(1210.4 m/s²)

F = 423.63 N

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