can you guys help me to solve first question????
Torque question

Answer:

Explanation:

Distance covered by sound = 2.7 m

speed of sound at 20⁰C = 343 m /s

time take by sound to cover the distance = distance / speed

= 2.7 / 343

= 7.87 ms . ( millisecond )

b )

Wavelength of sound = speed / frequency

= 343 / 523 m

= .6558 m

= 65.58 cm

Distance = 2.70 m = 270 cm

No of wavelengths contained in the distance

= 270 / 65.58

= 4.11 or 4 wavelengths ( by rounding off to digit )

The time taken by sound and the wavelengths of sound is required.

Time taken is [tex]0.00787\ \text{s}[/tex]

The number of wavelenghts is 4.

s = Distance = 2.7 m

v = Speed of sound at [tex]20^{\circ}\text{C}[/tex] = 343 m/s

Time is given by

[tex]t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{2.7}{343}\\\Rightarrow t=0.00787\ \text{s}[/tex]

f = Frequency = 523 Hz

Wavelength is given by

[tex]\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{343}{523}\\\Rightarrow \lambda=0.66\ \text{m}[/tex]

The wavelength is 0.66 m.

n = Number of wavelengths

[tex]n\lambda=s\\\Rightarrow n=\dfrac{s}{\lambda}\\\Rightarrow n=\dfrac{2.7}{0.66}\\\Rightarrow n\approx 4[/tex]

The number of wavelenghts is 4.

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Answer:

the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping is 3.3424 m

Explanation:  

Given the data in the question and as illustrated in the images below;

without the ladder immediately slipping, the net torque and the net force mus all balance out.

from the first image;

In the x, the force is;

[tex]F_{f}[/tex] = N₂

mg = N₁

the torque about the ground contact point gives the following equation

N₂Lsin∅ = mgcos∅[tex]\frac{L}{2}[/tex]

solving for ∅

tan∅ = mg / 2N₂      55      

∅ = tan⁻¹ (  mg / 2N₂ )    

we already know that N₂ = [tex]F_{f}[/tex]  = μN₁ = μmg

so,

∅ = tan⁻¹ (  mg / 2μmg )

∅ = tan⁻¹ (  1 / 2μ )

given that; The coefficient of static friction between the level ground and the foot of the ladder μ = 0.35

we substitute

∅ = tan⁻¹ (  1 / (2×0.35 ) )

∅ = tan⁻¹ ( 1.42857 )

∅ = 55°

now to get the required distance;

from the second image; cos∅ = d / L

d = Lcos∅

given that; length of the ladder = 5.2 m

we substitute

d = 5.2cos(50)

d = 3.3424 m

Therefore, the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping is 3.3424 m

Answer:

513 m

Explanation:

We have;

final speed of the airplane = 193.0 km/h * 1000/3600 = 53.6 m/s

acceleration of the air plane = 2.80 m/s2

initial velocity of the airplane = 0 m/s

length of the runway = distance covered

v^2 = u^2 + 2as

v^2 - u^2 = 2as

s = v^2 - u^2/2a

s = (53.6)^2 - 0^2/ 2 *  2.80

s = 2872.96/ 5.6

s = 513 m

Answer: B>A=D>C

Explanation:

Kinetic Energy is the product of mass and square of the velocity

For Jogger A

[tex]K.E._a=\frac{1}{2}mv^2[/tex]

For Jogger B

[tex]K.E._b=\frac{1}{2}\times\frac{m}{2}\times(3v)^2=\frac{9}{4}mv^2\\K.E._b=2.25mv^2[/tex]

For Jogger C

[tex]K.E._c=\frac{1}{2}\times3m\times(\frac{v}{2})^2=\frac{3}{8}mv^2\\K.E._c=0.375mv^2[/tex]

For Jogger D

[tex]K.E._d=\frac{1}{2}\times4m\times(\frac{v}{2})^2=\frac{1}{2}mv^2[/tex]

Kinetic Energy of Joggers in increasing order

B>A=D>C

Answer:

K_{e} = 2.0 J

Explanation:

In this exercise you are asked to calculate the elastic potential energy of a spring

          [tex]K_{e}[/tex] = ½ k x²

where k is the spring constant and x is the displacement from equilibrium position

In this exercise, indicate that the spring constant is k = 100 N/m, the length at rest is  x₀ = 20 cm = 0.20 m, up to the position x₁ = 40 cm = 0.40 m, therefore the elongation

           Δx = x₁ - x₀

           Δx = 0.40 - 0.20

           Δx = 0.20 m

let's calculate the elastic potential energy

           K_{e} = ½ 100 0.20²

           K_{e} = 2.0 J

Answer:

Nitrogen and Oxygen are the two major atmospheric gases.

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

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

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

[tex](1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s][/tex]

Answer:

    I = 1.06886  N s

Explanation:

The expression for momentum is

          I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

          sin 28.2 = v_y / v

          cos 28.2=  vₓ / v

          v_y = v sin 282

          vₓ = v cos 28.2

          v_y = 42.8 sin 28.2 = 20.225 m / s

          vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

         v_y = -20.55 m / s

         v_x = 37.72 m / s

X axis

         Iₓ = Δpₓ = [tex]p_{fx} - p_{ox}[/tex]

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

         Δpₓ = m [tex]v_{fx}[/tex] - m v₀ₓ = 0

          v_{fx} = v₀ₓ

therefore

          Iₓ = 0

Y axis  

        I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

         v_{fy} = - v_{oy}

         Δp_y = 2 m v_{oy}

         Δp_y = 2 0.0260 (20.55)

         [tex]\Delta p_{y}[/tex] = 1.0686 N s

the total impulse is

          I = Iₓ i ^ + I_y j ^

          I = 1.06886  j^ N s

Answer:

Angular momentum of the system is 16221465.4617 kgm²/s

Explanation:

Given that;

length of the side of the triangle L =  82 M

m = 75.0 kg × 100 = 7500 kg

distance of each vertex from center R = L/√3 = 82/√3 = 47.34 m

effective acceleration a = 9.8 / 2 = 4.9 m/s²

we know that; effective acceleration is being provided by centripetal acceleration.

so

a = R × w²

rate of rotation w = √( a / R) = √( 4.9 / 47.34)  = 0.3217 rad/seconds

Moment of Inertia I = 3mR²

we substitute

I = 3 × 7500 × (47.34)²

Also, Angular momentum L is expressed as;

L = I × w

so

L = 3 × 7500 × (47.34)² × 0.3217

L =  16221465.4617 kgm²/s

Therefore, Angular momentum of the system is 16221465.4617 kgm²/s

The Angular momentum of the system is 16221465.4617 kgm²/s

Given that;

Length of the side of the triangle L =  82 M

[tex]m = 75.0 kg \times 100 = 7500 kg[/tex]

Now

distance of each vertex from center R = L/√3

= 82/√3

= 47.34 m

Now

effective acceleration [tex]a = 9.8 \div 2[/tex] = 4.9 m/s²

So,

a = R × w²

Now

rate of rotation [tex]w = \sqrt( a \div R) = \sqrt( 4.9 \div 47.34)[/tex]  = 0.3217 rad/seconds

Moment of Inertia I = 3mR²

Now

[tex]I = 3 \times 7500 \times (47.34)^2[/tex]

Also, Angular momentum L is expressed as;

L = I × w

so

L = 3 × 7500 × (47.34)² × 0.3217

=  16221465.4617 kgm²/s

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Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q  / ε₀

E = Q  / 4πε₀r²

E ∝ 1 / r² .

Answer:

h = 31.9 m

Explanation:

When the ball travels from the ground to the highest point, its kinetic energy is converted into potential energy. So by the law of conservation of energy:

[tex]Kinetic\ Energy\ Lost\ by\ ball = Potential\ Energy\ gained\ by\ ball\\\frac{1}{2}mv^{2} = mgh\\\\h = \frac{v^{2}}{2g}[/tex]

where,

h = height gained by ball = ?

v = speed of ball = 25 m/s

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(25\ m/s)^{2}}{2(9.8\ m/s^{2})}[/tex]

h = 31.9 m

Answer:

33.5J

Explanation:

Given:

Mass of the canister= 3.1 kg

Initial velocity of canister= v(i)= 4.4i m/s

Final velocity of canister= v(f)= 6.4j m/s

Force magnitude( xy plane)= 5 N

The magnitude of vector V'= Vxi + Vyj + Vzk

|V|= √( Vx^2 + Vy^2 + Vz^2

From Kinectic energy and work theorem.

Net work = Kinectic energy of the canister

ΔK= W

(Kf - Ki)= W

Where Kf= final Kinectic energy

= 1/2 mv^2

If we input the given values we have,

= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J

Ki= initial Kinectic energy

= 1/2 mv^2

If we substitute the given values we have

=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J

Work done by canister = (final Kinectic energy - initial energy)

= 63.5- 30

=33.5J

Hence, work done on the canister 33.5J

Answer:

the simplified expression is written as 3.4 x 10³

Explanation:

Given expression;

[tex]\frac{34000\times 0.4}{0.02 \times 200}[/tex]

in scientific notation, the expression is simplified as;

[tex]\frac{34000\times 0.4}{0.02 \times 200} = \frac{13600}{4} = 3400 = 3.4 \times 10^3[/tex]

Therefore, in scientific notation, the simplified expression is written as 3.4 x 10³

Answer:

6.67A

Explanation:

The voltage across the capacitor formula is expressed as;

VL = IXL

VL is the voltage across the capacitor = 240volts (since it is a parallel connection, all the elements will have the same voltage)

I is the capacitor current

XL is the capacitive reactance = 36 ohms

Recall from the formula:

VL = IXL

I = VL/XL

I = 240/36

I = 6.67A

Hence the capacitor current is 6.67A

Answer:

D, 1J

Explanation:

PE=1/2kx^2 and plug in the variables.

PE=1/2 x 200 x 0.1^2= 1J

Answer:

The minibus traveled 75.66 km

Explanation:

Motion with Constant Speed

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

The formula to calculate the speed is:

[tex]\displaystyle v=\frac{d}{t}[/tex]

Where  

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we solve for d:

d = v . t

The minibus has a constant speed of v=39 km/h and it's required to find the distance it travels in t=1.94 hours.

Calculating the distance:

d = 39 km/h * 1.94 h

d = 75.66 km

The minibus traveled 75.66 km

Answer:

The direction could change

Answer:

14.8m

Explanation:

Given parameters:

Initial speed  = 17m/s

Unknown:

Maximum height  = ?

Solution:

At the maximum height, the final speed will be 0m/s;

 We use of the kinematics equation to solve this problem.

     V²   = U²   - 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height

         0²   = 17²  -  (2  x 9.8 x h )

          0 = 289  - (9.6h)

         -289  = -19.6h

            h  = 14.8m

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

[tex]F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})[/tex] (1)

Where:

[tex]F[/tex] - Explosive force, measured in newtons.

[tex]\Delta s[/tex] - Barrel length, measured in meters.

[tex]m[/tex] - Mass of the shell, measured in kilograms.

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the shell, measured in meters per second.

If we know that [tex]m = 1250\,kg[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 750\,\frac{m}{s}[/tex] and [tex]\Delta s = 15\,m[/tex], then the explosive force experienced by the shell inside the barrel is:

[tex]F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}[/tex]

[tex]F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}[/tex]

[tex]F = 23437500\,N[/tex]

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Answer:

[tex]D=99.4665307m \approx 99.5m[/tex]

Explanation:

From the question we are told that

Mass  [tex]m=6kg[/tex]

Velocity of mass  [tex]V_m=16[/tex]

Force of Tunnel  [tex]F_t=8N[/tex]

Length of Tunnel [tex]L_t=1.6[/tex]

Height of frictional incline [tex]H_i=2.9[/tex]

Angle of inclination  [tex]\angle =16 \textdegree[/tex]

Acceleration due to gravity  [tex]g=9.8m/s^2[/tex]

First Frictional surface has a coefficient  [tex]\alpha_1 =0.21\ for\ d_c=1[/tex]

Second Frictional surface has a coefficient [tex]\alpha _2=0.1[/tex]

Generally the initial Kinetic energy is mathematically given by

[tex]K.E=\frac{1}{2}mv^2[/tex]

[tex]K.E=\frac{1}{2}(6)(16)^2[/tex]

[tex]K.E=768[/tex]

Generally the work done by the Tunnel is mathematically given as

[tex]w_t=F_t*d_t[/tex]

[tex]w_t=8*1.6[/tex]

[tex]w_t=12.8J[/tex]

Therefore

[tex]Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t[/tex]

[tex]E_t=K.E+E_t\\E_t=768J+12.8J[/tex]

[tex]E_t=780.8J[/tex]

Generally the energy lost while climbing is mathematically given as

[tex]E_c=mgh[/tex]

[tex]E_c=(6)(9.8)(2.9)[/tex]

[tex]E_c=170.52J[/tex]

Generally the energy lost to friction is mathematically given as

[tex]E_f=\alpha *m*g*cos\textdegree*d_c[/tex]

[tex]E_f=0.21*6*9.8*cos16*1[/tex]

[tex]E_f=11.86965942 \approx 12J[/tex]

Generally the energy left in the form of mass [tex]Em[/tex] is mathematically given as

[tex]E_m=E_t+E_c+E_f[/tex]

[tex]E_m=(768J)-(170.52)-(12)[/tex]

[tex]E_m=585.48J[/tex]

Since

[tex]E_m=\alpha_2*g*m*d[/tex]

Therefore

It slide along the second frictional region

[tex]D=\frac{585.46}{0.1*9.81*6}[/tex]

[tex]D=99.4665307m \approx 99.5m[/tex]

Answer:

Double helix

Explanation:

The Double helix is a DNA molecule. The two strands around the Double Helix is called the twisted ladder.

Answer:

double helix

Explanation:

Answer:

The speed (magnitude of the velocity) is 14.4 m/s

Explanation:

Vertical Launch Upwards

It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the speed vf at any time is calculated by:

[tex]v_f=v_o-g.t[/tex]

A tennis ball is launched vertically up with an initial speed of vo=34 m/s. At time t=2 s, its speed is:

[tex]v_f=34-9.8*2[/tex]

[tex]v_f=34-19.6[/tex]

[tex]v_f=14.4\ m/s[/tex]

The speed (magnitude of the velocity) is 14.4 m/s

Beam balance is used to determine the mass of an object.

Answer:

a) phere A has 8 negative charges and sphere B has 2 positive charges

b)  each one has then 3 negative charges

Explanation:

In this case, it is asked to determine the charge of the spheres in two conditions

a) Before connecting the cable, sphere A has 8 negative charges and sphere B has 2 positive charges

b) After connecting the cable, as the spheres are metallic, the load is distributed, we have

              q = 8 -2 = 6 negative charges

these charges are distributed between the two spheres, each one has then 3 negative charges

Answer:

1- 4Ω

Explanation:

if 2Ω is 2 meters, than 4 metres is 4Ω