Check all the boxes that correspond to forces doing zero work
gravity acting on a projectile that has been launched straight up
You pushing your stalled car, but it doesn't budge
Drag acting on a plane flying horizontally
gravity acting on your car as it drives horizontally
An olympic weightlifter holding a barbell above their head (the barbell is motionless)
sliding friciton acting on a car slilding to a stop

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

Answer:

8kg

Explanation:

For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :

M ×3 = 12 ×2

M = 24/3 = 8kg

Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.

Answer:

Element C will be best for a nuclear fission reaction

Explanation:

Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process. For the reaction to be continuous in a chain reaction, the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.

Answer:

The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons

Explanation:

Binding energy can be given by the formula:

Binding energy = Binding energy per nucleon * Number of nucleons

This means that if the binding energy per nucleon = x MeV

Where x is a positive real number

The nucleus with 60 nucleons will have Binding energy = 60x MeV

The nucleus with 200 nucleons will have binding energy = 200x MeV

For a +ve x, 200x > 60x

Binding energy is proportional (directly) to nucleon volume. A further explanation is provided below.

Binding energy involving 200 nucleons would've been greater than 60 nucleons because so many more nucleons result in what seems like a stronger reaction.

It makes absolutely no difference out whether nucleon seems to be a proton as well as a neutron because they just have a similar strong coupling relatively steady.

Thus the response above is correct.

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Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

[tex]M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}[/tex]

For N vector:

[tex]N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}[/tex]

The resultant vector is the sum of the components of M and N:

[tex]F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}[/tex]

The magnitude of the resultant vector is:

[tex]|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm[/tex]

And the direction of the vector is:

[tex]\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°[/tex]

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Answer:

B) The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.

Explanation:

For every negative charge of an electron, there is an equal positively charged proton in the nucleus of the atom. This is why the overall charge on an atom is zero.

Answer:

B

Explanation:

This is described by Gauss a scientist.

The positive charge is found in the proton in the nucleus.

The neutron has no charge.

The positive charge radiates in all directions and a counter negative charge ensues.

Answer:

J = 8.4 kg*m/s

Explanation:

The magnitude of the impulse, J, on the ball can be calculated using the following equation:

[tex] J = F*t [/tex]   (1)

Where:

F: is the force = 12000 N

t: is the time = 0.70x10⁻³ s

So, by entering the values above into equation (1) we can find the impulse on the ball:

[tex] J = F*t = 12000 N*0.70 \cdot 10^{-3} s = 8.4 kg*m/s [/tex]

Therefore, the impulse on the ball is 8.4 kg*m/s.

I hope it helps you!

The magnitude J of the impulse on the ball is 8.4 kg m/s.

Since it applies a force of 12,000 N to the ball for a time of 0.70×10−3s.

So,

The magnitude is

[tex]= Force \times time\\\\= 12,000 \times 0.70\times 10^{-3}[/tex]

= 8.4 kg m/s

Here we basically multiplied the force with the time to determine the magnitude.

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Answer:

185 N

Explanation:

Sum of forces in the x direction:

Fₓ = -(80 N cos 75°) + (120 N cos 60°)

Fₓ = 39.3 N

Sum of forces in the y direction:

Fᵧ = (80 N sin 75°) + (120 N sin 60°)

Fᵧ = 181.2 N

The magnitude of the net force is:

F = √(Fₓ² + Fᵧ²)

F = √((39.3 N)² + (181.2 N)²)

F = 185 N

We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that  the magnitude of the resultant force is

R=200N

From the question we are told

Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?

A) 47.5 N

B) 185 N

C) 198 N

D) 200 N

Generally the equation for the Resultant force is mathematically given as

For x axis resolution

[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]

For y axis resolution

[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]

Therefore

[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]

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Answer:

605 J

Explanation:

Efficiency = work out / work in

0.78 = W / 775 J

W = 605 J

Answer:

Assuming that the vertical speed of the ball is 14 m/s we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:  

[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]

Where:

[tex]V_{f}[/tex]: is the final speed = 14 m/s

[tex]V_{0}[/tex]: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]  

b) The maximum height is:

[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]

[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]

c) The time can be found using the following equation:

[tex] V_{f} = V_{0} - gt [/tex]

[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]

d) The flight time is given by:

[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]

I hope it helps you!    

Answer:

The coefficient of static friction is 0.26

Explanation:

Given;

radius of the road, R = 97 m

banking velocity, V₁ = 75 km/h = 20.83 m/s

velocity of the moving car, V₂ = 100 km/h = 27.78 m/s

Car in a banked circular turn:

[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} )[/tex]

where;

θ is the angle between the horizontal ground and road in which the car move on

[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} ) \\\\\theta = tan^{-1}(\frac{20.83^2}{9.8*97} ) \\\\\theta = 24.5^o[/tex]

During this type of motion, the body acquires some acceleration which tends to retain the circular motion towards its center, known as centripetal acceleration.

There are two components of this acceleration;

Parallel acceleration,  [tex]a_|_|[/tex] = a*Cosθ

Perpendicular acceleration, a⊥ = a * Sinθ

Parallel acceleration, [tex]a_|_|[/tex]  [tex]= \frac{V^2*Cos \theta}{R}[/tex]

Perpendicular acceleration, a⊥ [tex]= \frac{V^2*Sin \theta}{R}[/tex]

Apply Newton's second law of motion;

sum of perpendicular forces acting on the car;

ma⊥ [tex]= F_N - mg*cos \theta[/tex]

[tex]m(\frac{V^2*Sin \theta}{R} ) = F_N - mg*Cos \theta\\\\F_N = mg*Cos \theta + m(\frac{V^2*Sin \theta}{R} )[/tex] --------equation (1)

sum of parallel  forces acting on the car

m[tex]a_|_|[/tex] [tex]= mg*Sin \theta - F_s[/tex]

[tex]m(\frac{V^2*Cos \theta}{R} ) = mg*Sin \theta - F_s\\\\F_s = mg*Sin \theta - m(\frac{V^2*Cos \theta}{R} )[/tex] ---------equation (2)

Coefficient of static friction is given as;

[tex]\mu = \frac{F_s}{F_N}[/tex]

Thus, divide equation (2) by equation (1)

[tex]\frac{F_s}{F_N} = \frac{mg*Sin \theta - m(\frac{V^2*Cos \theta}{R}) }{mg*Cos \theta + m(\frac{V^2*Sin \theta}{R}) } \\\\\frac{F_s}{F_N} = \frac{g*Sin \theta - (\frac{V^2*Cos \theta}{R}) }{g*Cos \theta + (\frac{V^2*Sin \theta}{R}) }[/tex]

V = V₂ = 27.78 m/s

θ = 24.5°

R = 97 m

g = 9.8 m/s²

Substitute in these values and solve for μ

[tex]\frac{F_s}{F_N} = \frac{9.8*Sin(24.5)+ (\frac{27.78^2*Cos (24.5)}{97}) }{9.8*Cos (24.5) + (\frac{27.78^2*Sin (24.5)}{97}) }\\\\\frac{F_s}{F_N} = \frac{4.0641 \ - \ 7.2391}{8.91702 \ + \ 3.299} = -0.26\\\\| \mu| = 0.26[/tex]

Therefore, the coefficient of static friction is 0.26

Given that,

First charge = Q₁

Second charge = Q₂

The potential experienced by Q2 due to Q1 increases

We know that,

The electrostatic force between two charges is defined as

[tex]F=\dfrac{kQ_{1}Q_{2}}{r^2}[/tex]

Where,

k = electrostatic constant

[tex]Q_{1}[/tex]= first charge

[tex]Q_{2}[/tex]= second charge

r = distance

According to given data,

The potential experienced by Q₂ due to Q₁ increases.

We know that,

The potential is defined from coulomb's law

[tex]V=\dfrac{Q_{1}}{4\pi\epsilon_{0}r}[/tex]

[tex]V\propto\dfrac{1}{r}[/tex]

If r decrease then V will be increases.

If V decrease then r will be increases.

Since, V is increases then r will decreases that is moving Q₁ closer  to Q₂.

Hence, Moving Q₁ closer to Q₂.

(c) is correct option.

Answer:

36s

Explanation:

Let the objects be A and B.

Let the initial velocity of A be U and the initial velocity of B be 3U

The height sustain by A will be;

The final velocity would be zero

V2 = U2-2gH

Hence

0^2= U2 -2gH

H = U^2/2g

Similarly for object B, the height sustain is;

V2 = (3U)^2-2gH

Hence

0^2= 3U^2 -2gH

U2-2gH

Hence

0^2= U2 -2gH

H = 3U^2/2g

By comparism. The object with higher velocity sustains more height and so should fall longer than object A.

Now object A would take;

From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;

V=10×12=120m/s let g be 10m/S2

Similarly for object B,

The final velocity for B when it's falling it should be 3×that of A

Meaning

3V= gt

t =3V/g = 3× 120/10 = 36s

Answer:

[tex]\delta = 0.385\,m[/tex] (Compression)

Explanation:

The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:

[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]

Where:

[tex]P[/tex] - Load experimented by the bar, measured in newtons.

[tex]L[/tex] - Length of the bar, measured in meters.

[tex]A[/tex] - Cross section area of the bar, measured in square meters.

[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.

The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])

[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]

Where:

[tex]D_{o}[/tex] - Outer diameter, measured in meters.

[tex]D_{i}[/tex] - Inner diameter, measured in meters.

[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]

[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]

The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)

[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]

[tex]\delta = -3.852\times 10^{-4}\,m[/tex]

[tex]\delta = -0.385\,mm[/tex]

[tex]\delta = 0.385\,m[/tex] (Compression)

Answer:

i hope it will be useful for you

Explanation:

F=5.6×10^-10N

R=93cm=0.93m

let take m1 and m2 =m²

according to newton's law of universal gravitation

F=m1m2/r²

F=m²/r²

now we have to find masses

F×r²=m²

5.6×10^10N×0.93m=m²

5.208×10^-9=m²

taking square root on b.s

√5.208×10^-9=√m²

so the two masses are m1=7.2×10^-5

and m2=7.2×10^-5

Answer:

a)     d = (6.00 t i ^ + 0.500 t²) m , b)   v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d)  v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

       v = x / t

        x = v t

we substitute

        x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

         y = v₀ t + ½ a t²

         y = ½ at t²

we substitute

        y = ½ 1.00 t²

        y = 0.500 t²

in vector position is

       d = x i ^ + y j ^

       d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

       vₓ = v₀

       vₓ = 6.00 m / s

y Axis  

there is an acceleration and the initial speed is zero

         [tex]v_{y}[/tex] = v₀ + a t

         v_{y} = a t

         v_{y} = 1.00 t

the velocity vector is

         v = vₓ i ^ + v_{y} j ^

         v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

        d = (6.00 4 i ^ + 0.50 4 2 j⁾

        d = (24.00 i ^ + 8.00 j^ ) m

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

        v = (6 i ^ + 1 5 j ^)

         v = (6.00 i ^ + 5 j^ ) m/s

Answer: Option A

Explanation:

The potential energy decreases in the case when the charges are opposite and they attract each other.

In this case there is no external energy required in order to put the charges together.

This is so because the charges are opposite and they will attract each other. Yes, the only condition should be that the charges should be alike.

Example: a negative charge and a positive charge.

Answer:

Explanation:

xf = xita + vxitb + ½axtc.

xf is displacement , dimensional formula L .

Xi initial displacement , dimensional formula L

t is time , dimensional formula T ,

vxi is velocity , dimensional formula LT⁻¹

ax is acceleration , dimensional formula = LT⁻²

xf = xi t a + vxi t b + ½ ax t c.

L = aLT + b LT⁻¹ T + c LT⁻² T

From the law of uniformity , dimensional formula of each term of RHS must be equal to term on LHS

aLT = L

a = T⁻¹

b LT⁻¹ T = L

b = 1 ( constant )

c LT⁻² T = L

c = T

so a = T⁻¹ , b = constant and c = T .

Answer:

98.73 m/s, 83.63 m/s, 87.9 m/s

Explanation:

From the question,

F = m(v-u)/t.............................. Equation 1

Where F = Force, m = mass, v = final velocity, u = initial velocity, t = time.

make u the subject of the equation

u = v-(Ft/m)......................... Equation 2

Given: F = 31  N, m = 11 kg, t = 4 s, v = 110 m/s

Substitute into equation 1

u = 110-(31×4/11)

u = 110-11.27

u = 98.73 m/s.

For, F = -10 N, v = 80 m/s

u = 80-(-10×4/11)

u = 80+3.63

u = 83.63 m/s

For, F = 47 N, v = 105 m/s

u = 105-(47×4/11)

u = 105-17.1

u = 87.9 m/s

The initial velocity at the the beginning of the time interval is 98.73 m/s, 83.63 m/s, 87.9 m/s

Answer:

The electric flux at the infinite surface is ZERO

Explanation:

From the question we are told that  

    The point charge are identical and the value is  [tex]q = 71.0 pC = 71 * 10^{-12} \ C[/tex]

    The distance of separation is  [tex]D = 29.0 \ cm = 0.29 \ m[/tex]

    The distance of both from the infinite surface is  d

Generally the electric force exerted by each of the  charge on the infinite surface is

       [tex]\phi = \frac{q}{\epsilon_o}[/tex]

Now given from the question that they are identical, it then means that the electric flux of the first charge on the infinite surface will be nullified by the electric flux of the second charge hence the electric flux at that infinite surface due to this two identical charges is ZERO

Answer:

[tex]q=4\times 10^{-16}\ C[/tex]

Explanation:

It is given that,

The charge on an object is 2500 e.

We need to find how many coulombs in the object. The charge remains quantized. It says that :

q = ne

[tex]q=2500\times 1.6\times 10^{-19}\ C\\\\q=4\times 10^{-16}\ C[/tex]

So, the charge on the object is [tex]4\times 10^{-16}\ C[/tex].

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex]

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

[tex]\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}[/tex]

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

[tex]Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C[/tex]

Finally, you obtain for E:

[tex]E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}[/tex]

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

Answer:

# _units = 1000

Explanation:

This exercise we can use a direct proportion rule.

If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?

    # _units = V₁₀ / V₁

The volume of a body of radius 1 is

       V₁ = 4/3 π r₁³

        V₁ = 4/3π

the volume of a body of radius r = 10

        V₁₀ = 4/3 π r₂³

        V10 = 4/3 π 10³

the number of times this content is

         #_units = 4/3 π 1000 / (4/3 π 1)

        # _units = 1000

Answer:

Seismic waves

Explanation:

seismic waves are not represented by electromagnetic graphs, nor can they be reflected on an electromagnetic spectrum. Visible light, radio waves, and microwaves are all electromagnetic waves, which are represented by graphs and electromagnetic spectrums.

Answer:

Siesmic waves

Explanation:

Answer:

Explanation:

We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth

[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]

[tex]L=\frac{.5}{.24}[/tex]

L= 2.05

The length will appear to be 2.05 m . and width will appear to be .5 m  to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.  

Answer:

Explanation:

a )

Iz( t ) = A + B t²

Iz( t ) = angular velocity

putting dimensional formula

T⁻¹ = A + Bt²

A = T⁻¹

unit of A is rad s⁻¹

BT² = T⁻¹

B = T⁻³

unit of B is rad s⁻³

b )

Iz( t ) = A + B t²

dIz / dt = 2Bt

angular acceleration = 2Bt

at t = 0

angular acceleration = 0

at t = 5

angular acceleration = 2 x 1.6 x 5

= 16 rad / s²

Iz( t ) = A + B t²

dθ / dt = A + B t²

integrating ,

θ = At + B t³ / 3

when t = 0 , θ = 0

when t = 2

θ = At + B t³ / 3

= 2.65 x 2 + 1.6 x 2³ / 3

= 5.3 + 4.27

= 9.57 rad .

Flywheel turns by 9.57 rad during first 2 s .

(a) The unit of A is rad/s and the unit of B is rad/s³.

(b) The angular acceleration of the flywheel at 0 s is 0 and at 5 s is 16 rad/s²

(c) The angular displacement of the flywheel during the first 2 seconds, is 9.57 rad.

The given parameters;

The units of A and B if z(t) is in radian per second (rad/s), are calculated as follows;

[tex]z(t)[\frac{rad}{s} ] = A[\frac{rad}{s} ] \ + \ Bt^2[\frac{rad}{s^3} ][/tex]

Thus, the unit of A is rad/s and the unit of B is rad/s³.

The angular acceleration of the flywheel is calculated as follows;

[tex]a = \frac{d\omega }{dt} =2Bt[/tex]

when, t = 0

a = 2(1.6)(0) = 0

when t = 5 s

a = 2(1.6)(5) = 16 rad/s²

The angular displacement of the flywheel during the first 2 seconds, is calculated as follows;

[tex]\theta = \int\limits {z(t)} \, dt\\\\\theta = At \ + \ \frac{Bt^3}{3} \\\\when, \ t = 2\ s;\\\\\theta = (2.65\times 2) \ + \ (\frac{1.6\times 2^3}{3} )\\\\\theta = 9.57 \ rad[/tex]

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Explanation:

impulse J = m × (v2-v1) =750 × ( 5 - 0 ) =3750( N×s)

Answer:

3750Ns

Explanation:

Impulse is defined as Force × time

Force = mass × acceleration,

Hence impulse is;

mass × acceleration × time.

From Newton's second law

Force × time = mass × ∆velocity

750× 5 = 3750Ns

∆velocity = Vfinal-Vinitial ; the initial velocity is zero since the body starts from rest.

Answer:

4.3 m/s

Explanation: