Calculate the pressure exerted on the ground when a woman wears high heals. Her mass is 65 kg. and the area of each heal is 1 cm^2.

Answer:

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Explanation:

Work done at fixed pressure, W = PΔV

Pressure, P = 3.7 x 10⁵ Pa

Change in volume, ΔV = 1.6 m³

Substituting the values of pressure and change in volume we will get

Work done at fixed pressure, W = PΔV =  3.7 x 10⁵ x 1.6 = 5.92 x 10⁵ J

Work is done by the gas  = 5.92 x 10⁵ J = 592 kJ

Answer:

The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

Explanation:

Given that,

Mass = 2 kg

Radius = 0.5 m

Angular speed = 3 rad/s

Force = 10 N

(I). We need to calculate the rotational kinetic energy

Using formula of kinetic energy

[tex]K.E =\dfrac{1}{2}\timesI\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}\times mr^2\times\omega^2[/tex]

[tex]K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2[/tex]

[tex]K.E=2.25\ J[/tex]

(II). We need to calculate the instantaneous change rate of the kinetic energy

Using formula of kinetic energy

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

On differentiating

[tex]\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}[/tex]

[tex]\dfrac{K.T}{dt}=mva[/tex]....(I)

Using newton's second law

[tex]F = ma[/tex]

[tex]a= \dfrac{F}{m}[/tex]

[tex]a=\dfrac{10}{2}[/tex]

[tex]a=5 m/s^2[/tex]

Put the value of a in equation (I)

[tex]\dfrac{K.E}{dt}=mva[/tex]

[tex]\dfrac{K.E}{dt}=mr\omega a[/tex]

[tex]\dfrac{K.E}{dt}=2\times0.5\times3\times5[/tex]

[tex]\dfrac{K.E}{dt}=15\ J/s[/tex]

Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

The rotational kinetic energy of the hoop is 2.25 J, and the instantaneous change rate of the kinetic energy is 15 W.

(a) To determine the rotational kinetic energy (RKE) of the hoop, we use the formula for RKE, which is given by:

[tex]\[ RKE = \frac{1}{2} I \omega^2 \][/tex]

where I is the moment of inertia of the hoop and [tex]\( \omega \)[/tex] is the angular speed. The moment of inertia I for a hoop (or a thin-walled cylinder) rotating about its center is:

[tex]\[ I = m r^2 \][/tex]

where m is the mass of the hoop and r is the radius. Given that  m = 2kg and ( r = 0.5 ) m, we can calculate I:

[tex]\[ I = 2 \times (0.5)^2 = 0.5 \text{ kg} \cdot \text{m}^2 \][/tex]

Now, we know [tex]\( \omega = 3 \)[/tex] rad/s. We can substitute the values into the RKE formula:

[tex]\[ RKE = \frac{1}{2} \times 0.5 \times (3)^2 = \frac{1}{2} \times 0.5 \times 9 = 2.25 \text{ J} \][/tex]

So, the rotational kinetic energy of the hoop is 2.25 J.

(b) The instantaneous change rate of the kinetic energy is equal to the power input to the system. Power P due to a torque [tex]\( \tau \)[/tex] is given by:

[tex]\[ P = \tau \omega \][/tex]

The torque [tex]\( \tau \)[/tex] can be calculated from the force F applied tangentially at the rim of the hoop:

[tex]\[ \tau = F \times r \][/tex]

Given that ( F = 10 ) N and ( r = 0.5 ) m, we find:

[tex]\[ \tau = 10 \times 0.5 = 5 \text{ N} \cdot \text{m} \][/tex]

Now, we can calculate the power:

[tex]\[ P = \tau \omega = 5 \times 3 = 15 \text{ W} \][/tex]

Therefore, the instantaneous change rate of the kinetic energy is 15 W.

Answer:

75 m/s is faster

Explanation:

MKS units stands for meter kilogram seconds

75 miles per hour = 75 mph

1 mile = 1609.34 meters

1 hour = 60×60 = 3600 seconds

1 mph = 1609.34/3600 = 0.44704 m/s

75 mph = 75×0.44707 = 33.52792 m/s

Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.

Answer:

Weight on Venus = 443.75 N

Explanation:

Weight of a body is the product of mass and acceleration due to gravity.

So we have

       Weight =  Mass x Acceleration due to gravity

        W = mg

        Mass, m = 50 kg

        Acceleration due to gravity, g = 8.875 m/s²

        W = 50 x 8.875 = 443.75 N

Weight on Venus = 443.75 N

Answer:

443.75 N

Explanation:

Weight is the force with which a planet can attract anybody towards its centre.

Weight = mass of body × acceleration due to gravity on that planet

Weight = 50 × 8.875 = 443.75 N

Answer:

Option (a), (b) and (c)

Explanation:

The resistance of a conductor depends on the length of the conductor, area of crossection of the conductor and the nature of the conductor.

The formula for the resistance is given by

R = ρ x l / A

Where, ρ is the resistivity of the conductor, l be the length of the conductor and A be the area of crossection of the conductor.

So, It depends on the length, area and the type of material.

Resistance of a material is determined by its length, cross-sectional area, and the type of material; voltage and current do not determine resistance but are influenced by it.

The Quantities Determining the Resistance of a Material

The resistance of a piece of material is determined by several key factors, specifically:

It is important to note that the voltage across the material and the current flowing through the material are not factors that determine resistance. These two quantities are actually influenced by the resistance according to Ohm's law, which states that Voltage (V) equals Current (I) times Resistance (R), or V = IR.

Answer:

164.33 N

Explanation:

Given:

The mass of the ball, m = 5.57 kg

Angle made by the sag = 19.4°

it is required to find the tension [tex]F_T[/tex] in the string

Note: Refer the attached figure

The tension in the string will be caused by the weight of the wall

thus,

Weight of the ball, W = 5.57 kg × 9.8 m/s² = 54.586 N

Now the resolving the tension [tex]F_T[/tex] in the string into components as shown in the figure attached, the weight of the ball will be balanced by the sin component of the tension

thus,

W = [tex]F_T[/tex]sinΘ

or

54.586 N = [tex]F_T[/tex] sin 19.4°

or

[tex]F_T[/tex] = 164.33 N

Hence, the tension in the string will be 164.33 N

To calculate the total resistance of three resistors connected in parallel, use the formula 1/R = 1/R1 + 1/R2 + 1/R3. The maximum error in the calculated value of R can be estimated by multiplying the sum of the errors in each resistance by the calculated value of R.

To find the total resistance of three resistors connected in parallel, we use the formula 1/R = 1/R1 + 1/R2 + 1/R3. Given the resistances R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, we can substitute these values into the formula to calculate the total resistance R. Therefore, 1/R = 1/100 + 1/25 + 1/10 = 0.01 + 0.04 + 0.1 = 0.15. Now, to estimate the maximum error in the calculated value of R, we consider the errors in each resistance. Since each resistance has a possible error of 0.5%, we can calculate the maximum error in R by multiplying the sum of the errors in each resistance by the calculated value of R. Therefore, maximum error in R = 0.005 * 0.15 = 0.00075 Ω.

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The maximum error in the total resistance R of three parallel resistors with a potential error of 0.5% in each resistor is approximately 0.667 ohms.

The question asks to calculate the maximum error in the calculated value of total resistance R when three resistors R1, R2, and R3 are connected in parallel, where R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, each with a possible error of 0.5%. The resistors in parallel have a total resistance denoted by:

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

To find the maximum error in the calculated value of R, we will first calculate R and then use derivative rules to estimate the maximum error considering the errors in R1, R2, and R3.

After calculating 1/R using the given resistances:

1/R = 1/100 + 1/25 + 1/10

1/R = 0.01 + 0.04 + 0.1 = 0.15

Therefore, R = 1 / 0.15 = 6.667 Ω

We calculate the maximum possible errors in resistances as:

Using the formula for the propagation of errors for functions of several independent variables, we estimate the maximum error in R (eR) as:

eR ≈ | -R² * eR1/R1² | + | -R² * eR2/R2² | + | -R² * eR3/R3² |

Plugging in the values:

eR ≈ | -6.667² * 0.5/100² | + | -6.667² * 0.125/25² | + | -6.667² * 0.05/10² |

eR ≈ | -0.04446 | + | -0.17784 | + | -0.4446 | = 0.667 Ω (Approximated to three decimal places)

The estimated maximum error in the calculated value of R is therefore approximately 0.667 Ω.

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Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure [tex]P=4.63\times10^{4}\ Pa[/tex]

Remove heat [tex]\Delta U= -1.95\times10^{4}\ J[/tex]

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

[tex]dU=Q-W[/tex]...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

[tex]dU=Q-PdV[/tex]

Where, W = PdV

Put the value in the equation

[tex]dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))[/tex]

[tex]dU=-17746.78\ J[/tex]

Hence, The change in internal energy of the system is -17746.78 J

Answer: 135 grams

Explanation:

[tex]Q_{absorbed}=Q_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]    

where,

[tex]m_1[/tex] = mass of ice = 100 g

[tex]m_2[/tex] = mass of aluminium cup =? g

[tex]T_{final}[/tex] = final temperature  =[tex]20^0C[/tex]

[tex]T_1[/tex] = temperature of ice = [tex]-10^oC[/tex]

[tex]T_2[/tex] = temperature of aluminium cup= [tex]70^oC[/tex]

[tex]c_1[/tex] = specific heat of ice= [tex]2.03J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of aluminium cup = [tex] 0.902 J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex][100\times 2.03\times (20-(-10))]=-[m_2\times 0.902\times (20-70)][/tex]

[tex]m_2=135g[/tex]

Therefore, the mass of the aluminium cup was 135 g.

By calculating the heat transfer between a 100 g ice cube and an aluminum cup, the mass of the cup is found to be approximately 978.3 g.

To solve this, we will use principles of thermal equilibrium and specific heat capacities. Let's go through the steps:

Step 1: Calculate the heat required to warm the ice to 0 degrees Celsius.

The specific heat capacity of ice is 2.1 J/g°C. The formula for heat is:

Q = m * c * ΔT

m = 100.0 g (mass of ice)

c = 2.1 J/g°C (specific heat capacity of ice)

ΔT = (0°C - (-10°C)) = 10°C

Q₁ = 100 g * 2.1 J/g°C * 10°C = 2100 J

Step 2: Calculate the heat required to melt the ice at 0 degrees Celsius.

The enthalpy of fusion of ice ,as we know ,is 334 J/g.

Q₂ = m * L

m = 100.0 g

L = 334 J/g

Q₂ = 100 g * 334 J/g = 33400 J

Step 3: Calculate the heat required to warm the melted ice from 0°C to 20°C.

The specific heat capacity of water is 4.18 J/g°C.

Q₃ = m * c * ΔT

m = 100.0 g

c = 4.18 J/g°C

ΔT = (20°C - 0°C) = 20°C

Q₃ = 100 g * 4.18 J/g°C * 20°C = 8360 J

Step 4: Calculate the total heat gained by the ice.

[tex]Q_{total[/tex] = Q₁ + Q₂ + Q₃ = 2100 J + 33400 J + 8360 J = 43860 J

Step 5: Calculate the heat lost by the aluminum cup.

The specific heat capacity of aluminum is 0.897 J/g°C. Since we need the heat lost, we use:

[tex]Q_{lost[/tex] = m * c * ΔT

[tex]Q_{lost[/tex] = 43860 J

c = 0.897 J/g°C

ΔT = (70°C - 20°C) = 50°C

Rearranging for m:

m = [tex]Q_{lost[/tex] / (c * ΔT)

m = 43860 J / (0.897 J/g°C * 50°C) = 978.3 g

Therefore, the mass of the aluminum cup is approximately 978.3 g.

Answer:

The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

Explanation:

Given that,

Electric field strength = 1500 N/C

We need to calculate the electric force

Using formula of electric field

[tex]F = Eq[/tex]

E = electric field strength

q = charge of electron

Electron has negative charge.

Put the value into the formula

[tex]F=1500\times(-1.6\times10^{-19})[/tex]

[tex]F=-2.4\times10^{-16}\ N[/tex]

Negative sign shows the opposite direction of the field

Hence, The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.

The magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

We know that electric force is given by the formula,

[tex]F = E \times q[/tex]

It is given that the electric field, E = 1500 N/C,

We also know that an electron is negatively charged and has a charge of 1.60217662 × 10⁻¹⁹ C.

[tex]F = E \times q\\\\F = 1500 \times 1.6 \times 10^{-19}\\\\F = 2.4 \times 10^{-16}\rm\ N[/tex]

Hence, the magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.

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Answer:

The magnetic field strength of the proton is 0.026 Tesla.

Explanation:

It is given that,

Speed of the proton, [tex]v=0.25\times 10^7\ m/s[/tex]

The radius of circular path, r = 0.975 m

It is moving perpendicular to a magnetic field such that the magnetic force is balancing the centripetal force.

[tex]qvB\ sin90=\dfrac{mv^2}{r}[/tex]

[tex]B=\dfrac{mv}{qr}[/tex]

q = charge on proton

[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.25\times 10^7\ m/s}{1.6\times 10^{-19}\ C\times 0.975\ m}[/tex]

B = 0.026 Tesla

So, the magnetic field strength of the proton is 0.026 Tesla.

Answer:

Depth of well 3.06m

Explanation:

We know that for a pipe closed at one end the frequencies are in ratios if 1:2:3:5:7.... to the fundamental frequency

In our case the given frequencies are in the ratio of

a)[tex]\frac{140}{196}=\frac{5}{7}[/tex]

b) [tex]\frac{196}{252}=\frac{7}{9}[/tex]

Thus the fundamental frequency can be calculated as [tex]140Hz=5n[/tex]

[tex]\therefore n=\frac{140}{5}=28Hz[/tex]

Now we know that

[tex]\lambda_{1}=4l\\\\\frac{v}{f}=4l\\\\l=\frac{v}{4f}[/tex]

Applying values we get

[tex]L=3.06m[/tex]

By identifying 56 Hz as the fundamental frequency and using the formula for the resonance of open-ended tubes, the depth of the well is calculated to be approximately 3.06 meters using the speed of sound as 343 m/s.

The student has observed standing waves at frequencies of 140 Hz, 196 Hz, and 252 Hz in a well. These frequencies represent the natural resonant frequencies or harmonics of the well. To find the depth of the well, we need to consider these frequencies as harmonics of a sound wave corresponding to the lengths of the air column in the well where each harmonic creates a standing wave.

Given the multiple frequencies, there is a constant difference of 56 Hz (196 Hz - 140 Hz, 252 Hz - 196 Hz) between consecutive frequencies. This indicates that 56 Hz is the fundamental frequency of the harmonics. We can represent these frequencies as [tex]F1 = 56 Hz \(n=1\), F2 = 112 Hz \(n=2\)[/tex]ere the observed frequencies correspond to F3, F4, and F5.

To determine the length of the well (L), we use the formula for the resonance of open-ended tubes, [tex]\(L = \frac{v}{2f}\)[/tex]d of sound (343 m/s) and 'f' is the frequency (56 Hz in this case). Therefore, [tex]L = \frac{343 m/s}{2 \times 56 Hz} = 3.0625 m[/tex]ly 3.06 meters.

Answer:

9.4 N

Explanation:

m = mass of the book = 1.83 kg

[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.522

[tex]\mu _{k}[/tex] = Coefficient of kinetic friction = 0.283

[tex]f_{s}[/tex] = Static frictional force

F = force needed to make the book move

force needed to make the book move is same as the magnitude of maximum static frictional force applied by the desk on the book

Static frictional force is given as

[tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

Hence, the force need to move the book is given as

F = [tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

F = [tex]\mu _{s}[/tex] mg

F = (0.522) (1.83 x 9.8)

F = 9.4 N

Answer:

The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω

Explanation:

Adding Resistances in serie is the way to build de largest equivalent value possible.

Rt= R1+R2+R3

Rt= 6.32 + 8.13 + 2.29

Rt= 16.74Ω

Answer:

a)W=20 KJ

b) ΔQ= 220 KJ

Explanation:

Given:

V₁=0.1 m^3,   P₁=200 kPa and heat is added to the system such that system expands under constant pressure.

Therefore V₂= 2V₁= 0.2 m^3

a) Work transfer W= P(V₂-V₁)= [tex]200\times(0.2-0.1)\times10^{5} = 2\times10^4 joules[/tex]

W=20 KJ

b) internal energy change ΔU= 200 KJ

from first law we know that ΔQ(net heat transfer)= ΔU + W

ΔQ= [tex]200\times10^3 +2\times10^4[/tex]

ΔQ=[tex]22\times10^4 J[/tex]

ΔQ= 220 KJ

Answer:

Worker's change in gravitational potential energy = 641.69 kJ

Explanation:

Potential energy = Mass x Acceleration due to gravity x Height

PE = mgh

Mass, m = 79 kg

Acceleration due to gravity, g = 9.81 m/s²

Height, h = 828 m

Potential energy, PE = 79 x 9.81 x 828 = 641691.72 J = 641.69 kJ

Worker's change in gravitational potential energy = 641.69 kJ

Answer:

(a) 4 rad/s^2

(b) 18 rad

Explanation:

w0 = 0, w = 12 rad/s, t = 3 s

(a) Let α be the angular acceleration.

w = w0 + α t

12 = 0 + 3 α

α = 4 rad/s^2

(b) Let θ be the angle rotated

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 4 x 9

θ = 18 rad

Answer:

2.13 h

Explanation:

Simple harmonic motion is:

x = A sin(2π/T t + φ) + B

where A is the amplitude, T is the period, φ is the phase shift, and B is the midline.

This can also be written in terms of cosine:

x = A cos(2π/T t + φ) + B

Here, A = d/2, T = 12.8, φ = 0, and B = d/2.  I'll use cosine so that the highest level is at t=0.

x = d/2 cos(2π/12.8 t) + d/2

When x = d - 0.250 d = 0.750 d:

0.750 d = d/2 cos(2π/12.8 t) + d/2

0.250 d = d/2 cos(2π/12.8 t)

0.500 = cos(2π/12.8 t)

π/3 = 2π/12.8 t

12.8/6 = t

t = 2.13

It takes 2.13 hours to fall 0.250 d from the highest level.

The time it takes for the water to fall a distance 0.250d from its highest level in the tide cycle can be calculated by determining a quarter of the period, which is 3.2 hours.

The time it takes for the water to fall a distance 0.250d from its highest level can be calculated using the concept of simple harmonic motion. Since the period of the tide is 12.8 hours, to find the time for the water to fall 0.250d, we need to determine the fraction of the period corresponding to this distance.

Given that the complete cycle from highest to lowest level is 12.8 hours, the time to fall 0.250d would be one-quarter of that period, which equals 3.2 hours.

Answer:

a)  The velocity of the ball upon leaving the foot = 600 m/s

b)  Time to reach the goal posts 40.0 m away = 0.07 seconds

c)  The kick won't e going inside goal post, it is higher by 10.34m.

Explanation:

a) Rate of change of momentum = Force

   [tex]\frac{\texttt{Final momentum - Initial momentum}}{\texttt{Time}}=\texttt{Force}\\\\\frac{0.040v-0.040\times 0}{0.010}=2400\\\\v=600m/s[/tex]

  The velocity of the ball upon leaving the foot = 600 m/s

b) Horizontal velocity = 600 cos20 = 563.82 m/s

   Horizontal displacement = 40 m

   Time

            [tex]t=\frac{40}{563.82}=0.07s[/tex]

   Time to reach the goal posts 40.0 m away = 0.07 seconds

c) Vertical velocity = 600 sin20 = 205.21 m/s

    Time to reach the goal posts 40.0 m away = 0.07 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 205.21 x 0.07 - 0.5 x 9.81 x 0.07²= 14.34 m

    Height of ball = 14.34 m

    Height of post = 4 m

    Difference in height = 14.34 - 4 = 10.34 m

    The kick won't e going inside goal post, it is higher by 10.34m.

180N

Using Newton's law of motion;

∑F = m x a       --------------------(i)

Where;

∑F = Resultant force

m = mass of the object (sled in this case)

a = acceleration of the sled

Calculate the resultant force;

Since the direction of motion is horizontal, the horizontal forces acting on the sled are the;

i. Applied force ([tex]F_{A}[/tex]) in one direction and;

ii. Frictional force ([tex]F_{R}[/tex]) in the other direction to oppose motion

Therefore, the resultant force ∑F is the vector sum of the two forces. i.e;

∑F = [tex]F_{A}[/tex] - [tex]F_{R}[/tex]  -----------------------(i)

Frictional force [tex]F_{R}[/tex] is the product of the coefficient of kinetic friction (μ) and weight(W) of the sled. i.e

[tex]F_{R}[/tex] = μ x W

Where;

W = mass(m) x gravity(g)

W = m x g

=> [tex]F_{R}[/tex] = μmg

Substitute [tex]F_{R}[/tex] into equation (ii)

∑F = [tex]F_{A}[/tex] - μmg

Substitute ∑F into equation (i)

[tex]F_{A}[/tex] - μmg = ma  -------------------(iii)

Since the motion is at constant speed, it means acceleration is zero (0)

Substitute a = 0 into equation (iii) to give;

[tex]F_{A}[/tex] - μmg = 0

=> [tex]F_{A}[/tex] = μmg

Substitute the values of μ = 0.3, m = 60kg and g = 10m/s² into the above equation to give;

=> [tex]F_{A}[/tex]  = 0.3 x 60 x 10

=> [tex]F_{A}[/tex] = 180N

This means that the applied force should be 180N

The amount of force the football player must apply to the sled is 176.4 Newton.

Given the following data:

We know that acceleration due to gravity (g) on Earth is equal to 9.8 [tex]m/s^2[/tex]

To find how much force the football player must apply to the sled:

Mathematically, the force of kinetic friction is given by the formula;

[tex]Fk = umg[/tex]

Where;

Substituting the given parameters into the formula, we have;

[tex]Fk = 0.30\times60\times9.8[/tex]

Force, Fk = 176.4 Newton.

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Answer:

The speed when it strikes the ground below is V= 157.64 m/s < -56.92º .

Explanation:

V= 150m/s

α= 55º

hi= 120m

Vy= V*sinα

Vy= 122.87 m/s

Vx= V * cos α

Vx= 86.03 m/s

h= hi + Vy * t - g*t²/2

clearing t we get the total flying time of the projectile:

t(total fly)= 26.01 sec

0= Vy - g*t

clearing t we get the maximum height time:

t(max height)= 12.53 sec

to get the fall time:

t(fall)= t(total fly) - t(max height)

t(fall)= 13.48 sec

Vy'= g* t(fall)

Vy'= 132.1 m/s

V'= √(Vx² +Vy'²)

V'= 157.64 m/s

α'= tg⁻¹ (Vy'/Vx)

α'= -56.92º

By using the conservation of energy principle, we find the speed of the projectile when it hits the ground to be roughly 165 m/s (rounded to 3 significant figures).

Given that a projectile is fired upward at an angle of 55° from the top of a 120 m cliff at a speed of 150 m/s, it is asked what its speed will be when it strikes the ground. To answer such a question, we employ the principle of conservation of energy, which states that the total mechanical energy (kinetic energy + potential energy) of an isolated system remains constant if non-conservative forces like air resistance are negligible.

Now, the energy of the projectile at the top of the cliff is equal to its kinetic energy (as it's launched) and its potential energy (due to its height). When the projectile hits the ground, all its potential energy will be converted into kinetic energy as the body has descended from the height, which the body will possess as its speed.

Let's calculate:

Initial Energy = Final Energy

1/2 * mass * (initial speed)² + mass * g * height = 1/2 * mass * (final speed)²

Solving for final speed, -(initial speed)² - 2gh = - (final speed)²

finalSpeed = sqrt((initial speed)² + 2*g*h)

Given that the initial speed = 150m/s, g (acceleration due to gravity) = 9.81m/s², and the height = 120m, the final speed would amount to 164.833 m/s. Therefore, when rounding to 3 significant figures, the speed when it hits the ground will be 165 m/s.

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Answer:

Part a)

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Explanation:

Part a)

Electrostatic force between two charged spherical balls is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we will have

[tex]q_1 = q_2 = 50 nC[/tex]

here the distance between the center of two balls is given as

[tex]r = 2.1 cm = 0.021 m[/tex]

now we will have

[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}[/tex]

[tex]F = 0.051 N[/tex]

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Answer:

Force of friction, F = 45.76 N

Explanation:

t is given that,

Mass of the skater, m = 64 kg

Initial velocity of the skater, u = 2.81 m/s

Finally it comes to rest, v = 0

Time, t = 3.93 s

We need to find the force of friction. According to seconds law of motion as :

F = m × a

[tex]F=m\times \dfrac{v-u}{t}[/tex]

[tex]F=64\ kg\times \dfrac{0-2.81\ m/s}{3.93\ s}[/tex]

F = −45.76 N

So, the frictional force exerting on the skater is 45.76 N. Hence, this is the required solution.

Answer:[tex]4.802 m/s^2[/tex]

Explanation:

height of slope(h) =90 foot

[tex]\theta =40[/tex]

weight of snowboarder=170 pounds\approx 77.1107 kg

[tex]\mu =0.2[/tex]

as the snowboarder is sliding down the slope therefore

Now net acceleration of snowboarder is

[tex]a_{net}=gsin\theta -\mu \cdot gcos\theta[/tex]

[tex]a_{net }=9.81\times sin\left ( 40\right )-0.2\times 9.81\times cos\left ( 40\right )[/tex]

[tex]a_{net }=4.802 m/s^2[/tex]

Answer:

1.5 m/s²

Explanation:

m = mass of the clarinet case = 3.450 kg

F = upward force on the clarinet through the air = 38.92 N

W = weight of the clarinet case in down direction

weight of the clarinet case is given as

W = mg

W = (3.450) (9.8)

W = 33.81 N

a = acceleration of the case

Force equation for the motion of the case is given as

F - W = ma

38.92 - 33.81 = (3.450) a

a = 1.5 m/s²

Explanation:

The motion of satellite is an example of uniform circular motion. In this type of motion, the velocity of object varies at each and every point but its speed is constant. The satellite in a circular orbit travel at a constant speed.

The gravitational force of earth is balanced by the centripetal force such that,

[tex]F_g=F_c[/tex]

A satellite travel at a constant speed because there is no force acting in the direction of motion of satellite i.e. there is no resistive force that opposite the motion of the satellite.

The average induced current in the loop is 0.218 A.

The emf induced in the loop is determined by applying Faraday's law as shown below;

emf = dФ/dt

emf = BA/t

where;

A is the area

A = πr² = πd²/4

A = π x (0.025)²/4

A = 4.908 x 10⁻³ m²

emf = (2 x 4.908 x 10⁻³)/(0.45)

emf = 2.18 x  10⁻³ V

The average induced current in the loop is calculated as follows;

I = emf/R

I = 2.18 x  10⁻³/0.01

I = 0.218 A

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Answer:

(a): The frequency of the waves is f= 0.16 Hz

Explanation:

T/4= 1.5 s

T= 6 sec

f= 1/T

f= 0.16 Hz (a)

(a) The frequency of the waves is ¹/₆ Hz ≈ 0.167 Hz

(b) The speed of the waves is 5¹/₃ m/s ≈ 5.33 m/s

[tex]\texttt{ }[/tex]

Let's recall the speed of wave and intensity of wave formula as follows:

[tex]\large {\boxed {v = \lambda f}}[/tex]

f = frequency of wave ( Hz )

v = speed of wave ( m/s )

λ = wavelength ( m )

[tex]\texttt{ }[/tex]

[tex]\large {\boxed {I = 2 \pi^2 A^2 f^2 \rho v}}[/tex]

I = intensity of wave ( W/m² )

A = amplitude of wave ( m )

f = frequeny of wave ( Hz )

ρ = density of medium ( kg/m³ )

v = speed of wave ( m/s )

Let's tackle the problem!

[tex]\texttt{ }[/tex]

Given:

time taken = t = 1.50 seconds

distance covered = d = 8.00 m

Asked:

(a) frequency of the waves = ?

(b) speed of the waves = ?

Solution:

[tex]t = \frac{1}{4}T[/tex]

[tex]1.50 = \frac{1}{4}T[/tex]

[tex]T = 4 \times 1.50[/tex]

[tex]T = 6 \texttt{ seconds}[/tex]

[tex]\texttt{ }[/tex]

[tex]f = \frac{1}{T}[/tex]

[tex]f = \frac{1}{6} \texttt{ Hz}[/tex]

[tex]\texttt{ }[/tex]

[tex]d = \frac{1}{4}\lambda[/tex]

[tex]8.00 = \frac{1}{4}\lambda[/tex]

[tex]\lambda = 8.00 \times 4[/tex]

[tex]\lambda = 32 \texttt{ m}[/tex]

[tex]\texttt{ }[/tex]

[tex]v = \lambda f[/tex]

[tex]v = 32 \times \frac{1}{6}[/tex]

[tex]v = 5\frac{1}{3} \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: College

Subject: Physics

Chapter: Sound Waves

Answer:

The contraction in the rod is 71 mm.

Explanation:

Given that,

original length L'= 2.99 m

Speed [tex]v= 6.49\times10^{7}\ m/s[/tex]

We need to calculate the length

Using expression for length contraction

[tex]L'=\gamma L[/tex]

[tex]L=\dfrac{L'}{\gamma}[/tex]

Where,

[tex]\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}[/tex]

[tex]L=\sqrt{1-\dfrac{v^2}{c^2}}L'[/tex]

Where, v = speed of observer

c = speed of the light

Put the value into the formula

[tex]L=\sqrt{1-\dfrac{(6.49\times10^{7})^2}{(3\times10^{8})^2}}\times2.99[/tex]

[tex]L=2.919\ m[/tex]

The expression for the contraction in the rod

[tex]d =L'-L[/tex]

[tex]d=2.99-2.919 [/tex]

[tex]d=0.071[/tex]

[tex]d= 71\ mm[/tex]

Hence, The contraction in the rod is 71 mm.

Answer:

The work done on the test charge is 4.5 joules

Explanation:

It is given that,

Electric potential difference, [tex]V=1000\ V[/tex]

Charged particle, [tex]q=4.5\times 10^{-3}\ C[/tex]

Work done on the test charge is given by the product of test charge and potential difference i.e.

W = q × V

[tex]W=4.5\times 10^{-3}\ C\times 1000\ V[/tex]

W = 4.5 Joules

So, the work done on the test charge is 4.5 joules. Hence, this is the required solution.