Which of the following objects has the greatest kinetic energy?A. a car traveling at 35 miles per hour B. a horse sprinting at 35 miles per hour C. a baseball flying at 35 miles per hour D. an insect flying at 35 miles per hour E. All four objects have the same kinetic energy.

Answer:

Explanation:

Arizona Nevada and California. That includes some pretty big cities. Los Angeles, Las Vegas, San Diego to name 3.

Answer:

The ball is in the air for 3.628 sec

Explanation:

We can find speed in the "y" direction when the ball lands with this equation:

[tex]V_{fy} ^{2} = V_{oy} ^{2} + 2*g*(y_{f} - y_{o})[/tex]

Where:   [tex]V_{oy} = 0 m/sec[/tex]   because the ball is thrown horizontally

              [tex]g = - 9.8 m/sec^{2}[/tex]   because gravity is a vector that points down in the "y" direction

              [tex]y_{o} = 64.5 m[/tex]

              [tex]y_{f} = 0 m[/tex]   because the ball lands on the ground

Then:      [tex]V_{fy} ^{2} = (0 m/sec)^{2} + 2(-9.8 m/sec^{2})(0 m - 64.5 m)[/tex]

              [tex]V_{fy} ^{2} = 2(- 9.8 m/sec^{2})(- 64.5 m)[/tex]

              [tex]V_{fy} ^{2} = 1264.2 m^{2}/sec^{2}[/tex]

              [tex]V_{fy} = \sqrt{1264.2 m^{2}/sec^{2}}[/tex]

              [tex]V_{fy} = - 35.5556 m/sec[/tex]   because the ball is falling and that means speed in the "y" direction is a vector that points down

Now we can calculate the time with next equation:

[tex]V_{fy} = V_{oy} + g*t[/tex]

[tex]V_{fy} - V_{oy} = g*t[/tex]

[tex]t = \frac{V_{fy} - V_{oy}}{g}[/tex]

Then:       [tex]t = \frac{-35.5556 m/sec - 0 m/sec}{- 9.8 m/sec^{2}}[/tex]

Finally      t = 3.628 sec

The ball is in the air for approximately 3.63 seconds.

Separate Motions: We can analyze the horizontal and vertical motions of the ball independently since air resistance is neglected.

Horizontal Motion: This is a constant velocity motion because there's no horizontal acceleration (ignoring air resistance).

Vertical Motion: This is a free-fall motion with constant acceleration due to gravity (g ≈ 9.81 m/s²).

Horizontal Distance (x):

Given: x = 103.8 m (horizontal distance from the building base)

Time (t): We need to find the total time (t) the ball spends in the air.

Vertical Motion:

Unknown: Time (t) for the vertical motion (which is the same as the total time in the air)

Known: Vertical distance (y) = -64.5 m (negative since downward) and acceleration due to gravity (g)

Vertical Kinematics Equation: We can use the kinematic equation for constant acceleration to solve for the time (t) for the vertical motion:

y = 1/2 * g * t² (We've plugged in the negative value for y since it's moving downward)

Solve for Time (t) in Vertical Motion:

Rearrange the equation to isolate t: t² = 2 * y / g

Substitute the known values: t² = 2 * (-64.5 m) / (9.81 m/s²) ≈ 13.19 s²

Take the square root of both sides to find t (remember there can be positive and negative time solutions, but we're interested in the positive time for the ball to be in the air).

t ≈ ± 3.63 s (We discard the negative solution)

However, this time represents only the vertical motion.

Relating Horizontal and Vertical Motion: Since the ball is thrown horizontally, the horizontal time (t_horizontal) is the same as the vertical time (t) we just calculated (t_horizontal = t_vertical).

Therefore, the ball is in the air for approximately 3.63 seconds.

Answer:

Answer to the question:

Explanation:

a. The pitch would get progressively lower (i.e., smaller frequency)

A person on the bridge would hear the pitch of an air horn during a bungee jump become lower as the jumper descends and higher as the jumper ascends, due to the Doppler effect.

During a bungee jump, if someone is blowing an air horn, a person on the bridge would experience a change in the perceived pitch of the sound due to the Doppler effect. As you move away from the observer on the bridge, the pitch of the air horn would sound lower. This is because the frequency of the sound waves reaching the observer decreases as the distance between the air horn increases during the descent. Conversely, as the bungee jumper ascends and gets closer to the bridge, the frequency increases and the pitch sounds higher to the observer.

The correct answer to the question is:

a. The pitch would get progressively lower (i.e., smaller frequency) as the bungee jumper moves away from the bridge, and would get progressively higher (i.e., larger frequency) as the jumper ascends back towards the bridge.

Answer: C. 8.0 m west

Explanation: The arrows are going 15 m west and 7.0 m east. 7 meters of the west will cancel out because 15-7=8. Subtract the smaller number from the bigger number, which is west minus east. The answer will be 8.0 m west.

A. The bullet's speed is [ ( M + m ) / m ] √ ( 2 μ g d )

B. The initial speed of the 9.0 g bullet is about 610 m/s

[tex]\texttt{ }[/tex]

Let's recall Impulse formula as follows:

[tex]\boxed {I = \Sigma F \times t}[/tex]

where:

I = impulse on the object ( kg m/s )

∑F = net force acting on object ( kg m /s² = Newton )

t = elapsed time ( s )

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

mass of bullet = m = 9.0 g = 9.0 × 10⁻³ kg

mass of block = M = 12 kg

sliding distance = d = 5.4 cm = 5.4 × 10⁻² m

coefficient of kinetic friction = k = 0.20

Asked:

initial bullet's speed = u₁ = ?

Solution:

Firstly, we will use Conservation of Energy formula to find the speed of the block:

[tex]W = \Delta Ek[/tex]

[tex]fd = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu N d = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu (M + m)g d = \frac{1}{2}(M+m)v^2[/tex]

[tex]\mu g d = \frac{1}{2} v^2[/tex]

[tex]\boxed {v = \sqrt{2 \mu g d}}[/tex] → Equation A

[tex]\texttt{ }[/tex]

Next, we will use Conservation of Momentum formula to find the initial speed of the bullet:

[tex]\texttt{Total Momentum Before Collision = Total Momentum After Collision}[/tex]

[tex]m u_1 + M u_2 = ( m + M ) v[/tex]

[tex]m u_1 + M (0) = ( m + M ) v[/tex]

[tex]m u_1 = ( m + M ) v[/tex]

[tex]m u_1 = ( m + M ) \sqrt { 2\mu g d}[/tex] ← Equation A

[tex]\boxed {u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}}[/tex]

[tex]\texttt{ }[/tex]

[tex]u_1 = \frac { m + M }{ m } \sqrt { 2\mu g d}[/tex]

[tex]u_1 = \frac { 9.0 \times 10^{-3} + 12 }{ 9.0 \times 10^{-3} } \sqrt { 2 \times 0.20 \times 9.80 \times 5.4 \times 10^{-2}}[/tex]

[tex]\boxed{u_1 \approx 610 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

(a) The expression for the speed of the bullet is [tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]

(b) The speed of the bullet at the given parameters is 613.78 m/s.

The given parameters;

Apply the principle of conservation of linear momentum to determine the initial speed of the bullet;

[tex]m_1 u_1 + m_2 u_2 = V(m_1 + m_2)\\\\m_1 u_1 + 0 = V(m_1 + m_2)\\\\u_1 = \frac{V(m_1 + m_2)}{m_1}[/tex]

Apply the principle of work-energy theorem to determine the speed of the bullet-block system;

[tex]K.E - P.E = W_f\\\\\frac{1}{2} MV^2 - 0 = \mu_k (Mg)d\\\\V^2 = 2\mu_k gd\\\\V = \sqrt{2\mu_k gd[/tex]

The expression for the speed of the bullet is written as;

[tex]u_1 = \frac{V(m_1 + m_2)}{m_1} \\\\u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1}[/tex]

The speed of the bullet at the given parameters is calculated as follows;

[tex]u_1 = \frac{(m_1 + m_2) \sqrt{2\mu_k gd} }{m_1} \\\\u_1 = \frac{(0.009 + 12) \sqrt{2\times 0.2 \times 9.8 \times 0.054} }{0.009} \\\\u_1 = 613.78 \ m/s[/tex]

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Answer:

Set the parking brake of the towing vehicle, and put it in park (or first gear if you have a manual transmission).

Move the vessel onto the trailer far enough to attach the winch line to the bow eye of the vessel. ...

Shut off the engine, and raise the engine or outdrive.

Pull the vessel out of the water.

Explanation:

The first thing you should do after retrieving a boat onto a trailer is to secure the boat to the trailer. This is important to ensure the boat remains stable and safe during transportation.

Here are the steps to secure the boat to the trailer:

1. Position the boat properly: Align the boat on the trailer so that it is centered and evenly distributed. Make sure the boat is positioned in such a way that the weight is balanced and evenly distributed across the trailer.

2. Attach the bow strap: The bow strap is a strong, adjustable strap that is used to secure the front (bow) of the boat to the trailer. Connect one end of the bow strap to the trailer and the other end to a secure point on the boat's bow. Make sure the strap is tight and secure, but not overly tightened to the point of damaging the boat.

3. Connect the stern tie-downs: Stern tie-downs are straps or ropes used to secure the rear (stern) of the boat to the trailer. Attach one end of each stern tie-down to the trailer and the other end to a secure point on the boat's stern. Make sure the stern tie-downs are tight and secure, but again, avoid over-tightening.

4. Check the connections: After attaching the bow strap and stern tie-downs, double-check all the connections to ensure they are properly secured. Give each strap a gentle tug to make sure it is tight and won't come loose during transport.

5. Secure loose items: Before hitting the road, secure any loose items in the boat, such as life jackets, paddles, or fishing gear. These items should be properly stowed and secured to prevent them from shifting or falling out during transportation.

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Answer:

The mass of the puck is 0.166 kg

Explanation:

It is given that,

Speed of the ice hockey puck, u = 12 m/s

Impulse delivered by the hockey stick, J = 4 kg-/s

After delivering an impulse, the puck move off in the opposite direction with the same speed, v = -12 m/s

We need to find the mass of the puck. The impulse is equal to the change in momentum i.e.

[tex]J=m(v-u)[/tex]

m is the mass of the puck

[tex]4=m(-12-12)[/tex]

[tex]4=-24\ m[/tex]

[tex]m=\dfrac{1}{6}=0.166\ kg[/tex]

So, the mass of the puck is 0.166 kg. Hence, this is the required solution.

When the mass of the puck that moves with the same speed in opposite direction is equal to 0.1667 kg.

Impulse is a large force that is applied to an object for a very short period of time. It is given by the formula,

[tex]J =m(v-u) = F(\delta t)[/tex]

where v and u are the final and initial velocity, F is the force that is been applied, m is the mass of the object, t is the time for which the force is been applied.

We know that the impulse can be written as,

[tex]J = m(v-u)[/tex]

Given to us

J  =  4 kg⋅m/s,

u = 12 m/s

v = -12 m/s(opposite in direction with the same speed)

Substitute the values,

[tex]4 = m(-12-12)\\\\4 = m(-24) \\\\m = -0.1666\\\\m = 0.1667 \rm\ kg[/tex]

Since the mass of the punk can not be negative, therefore, the mass of the puck is 0.1667 kg.

Hence, when the mass of the puck that moves with the same speed in opposite direction is equal to 0.1667 kg.

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Answer:

-99.1 V

Explanation:

For a free electron accelerated from rest, the final kinetic energy of the electron is equal to the change in electric potential energy of the electron:

[tex]K=\Delta U\\\frac{1}{2}mv^2 = q\Delta V[/tex]

where

[tex]m=9.11\cdot 10^{-31}kg[/tex] is the mass of the electron

[tex]v=5.9\cdot 10^6 m/s[/tex] is the final speed

[tex]q=-1.6\cdot 10^{-19} C[/tex] is the charge of the electron

[tex]\Delta V[/tex] is the potential difference

Solving for [tex]\Delta V[/tex], we find

[tex]\Delta V=\frac{mv^2}{2q}=\frac{(9.11\cdot 10^{-31})(5.9\cdot 10^6)^2}{2(-1.6\cdot 10^{-19})}=-99.1 V[/tex]

Answer:

Frequency [tex]f_{min,1}=155.90\ Hz[/tex]

Explanation:

Given data:

The distance between the speakers, d = 2.59 m

The distance between the listeners, ΔL = 22.7 - 21.6 = 1.1 m

Now, For a destructive interference, we know that

[tex]\frac{\Delta L}{\lambda}=0.5,1.5,2.5,.........[/tex]

where, λ = wavelength

thus,

frequency [tex]f_{min,n}=\frac{(n-0.5)v}{\Delta L}[/tex]

where,

v = speed of sound = 343 m/s

for n = 1

we get

frequency [tex]f_{min,1}=\frac{(1-0.5)\times 343}{1.1}[/tex]

or

Frequency [tex]f_{min,1}=155.90\ Hz[/tex]

Answer:

R = 964.5 m

Explanation:

When plane is moving in vertical loop then at the condition of free fall then force of gravity on the passengers will be balanced by the the pseudo force on them.

In ground frame we can say that normal force on the passengers will become zero

So we have

[tex]mg = \frac{mv^2}{r}[/tex]

[tex]m(9.8) = \frac{mv^2}{R}[/tex]

here we know that

v = 350 km/h = 97.22 m/s

now we have

[tex]9.8 = \frac{97.22^2}{R}[/tex]

[tex]R = 964.5 m[/tex]

Radius of curvature of the flight path must be approximately 966.67 meters to create a condition of apparent zero weight inside the aircraft cabin.

The condition of weightlessness inside an aircraft can be simulated when the centripetal acceleration equals the acceleration due to gravity. According to the formula a = v^2 / r, where a is the centripetal acceleration, v is the velocity, and r is the radius of curvature, we need to rearrange the formula to solve for r: r = v^2 / g. Given that the aircraft is moving at 350 km/h, which is approximately 97.22 m/s (since 1 km/h = 0.27778 m/s), and using g as 9.8 m/s^2, we can calculate the necessary radius of curvature to create a condition of weightlessness. Therefore, the radius of curvature r is:

r = (97.22 m/s)^2 / 9.8 m/s^2

r = 966.67 m

Thus, the radius of curvature of the flight path must be approximately 966.67 meters to create a condition of apparent zero weight inside the aircraft cabin.

Answer:

Part a)

[tex]W = (10 N)(2 m) = 20 J[/tex]

Part b)

[tex]\Delta K = 0.736 J[/tex]

Part c)

[tex]\Delta U = 13.2 J[/tex]

Part d)

[tex]U_{thermal} = 5.66 J[/tex]

Explanation:

Part a)

Work done by the applied force is given by the formula

[tex]W = F.d[/tex]

here we know that

[tex]F = 10 N[/tex]

[tex]d = 2 m[/tex]

[tex]W = (10 N)(2 m) = 20 J[/tex]

Part b)

As we know that the box was at rest initially and then it is moving with speed 0.80 m/s

so here we can say

[tex]\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]

[tex]\Delta K = \frac{1}{2}(2.3)(0.80)^2 - 0[/tex]

[tex]\Delta K = 0.736 J[/tex]

Part c)

Change in gravitational potential energy is given as

[tex]\Delta U = - W_g[/tex]

[tex]\Delta U = -(-mg sin\theta d)[/tex]

[tex]\Delta U = (2.3)(9.81)(sin17)(2)[/tex]

[tex]\Delta U = 13.2 J[/tex]

Part d)

Now by energy conservation law we can say that

Work done by external agent = change in kinetic energy + change in potential energy + thermal energy lost

so we have

[tex]20 = 13.6 + 0.736 + U_{thermal}[/tex]

[tex]U_{thermal} = 5.66 J[/tex]

Answer:

(a). The work done on the system by force is 20 J.

(b). The change in kinetic energy of the system is 0.736 J.

(c). The change in the gravitational potential energy of the system is 13.18 J

(d). The thermal energy of the system is 6.84 J.

Explanation:

Given that,

Mass of box = 2.3 kg

Force = 10 N

Length = 2.0 m

Angle = 17°

Speed = 0.80 m/s

(a). We need to calculate the work done

Using formula of work done

[tex]W=F\times d[/tex]

Put the value into the formula

[tex]W=10\times2.0[/tex]

[tex]W=20\ J[/tex]

The work done on the system by force is 20 J.

(b). We need to calculate the change in kinetic energy of the system

Using formula of change of kinetic energy

[tex]\Delta K.E=K.E_{f}-K.E_{i}[/tex]

[tex]\Delta K.E=\dfrac{1}{2}mv^2-0[/tex]

Put the value into the formula

[tex]\Delta K.E=\dfrac{1}{2}\times2.3\times(0.80)^2[/tex]

[tex]\Delta K.E=0.736\ J[/tex]

The change in kinetic energy of the system is 0.736 J.

(c). We need to calculate the change in the gravitational potential energy of the system

Using formula of gravitational potential energy

[tex]P.E=mgh\sin\theta[/tex]

Where, h = change in height

Put the value into the formula

[tex]P.E=2.3\times9.8\times2.0\sin17[/tex]

[tex]P.E=13.18\ J[/tex]

The change in the gravitational potential energy of the system is 13.18 J.

(d). We need to calculate the thermal energy of the system

Using formula of thermal energy

Work done=Change in kinetic energy+change in potential energy+change in thermal energy

[tex]\Delta U_{th}=W-\Delta K.E+-Delta P.E[/tex]

Put the value into the formula

[tex]\Delta U_{th}=20-0.736-13.18[/tex]

[tex]\Delta U_{th}=6.084\ J[/tex]

The thermal energy of the system is 6.84 J.

Hence, (a). The work done on the system by force is 20 J.

(b). The change in kinetic energy of the system is 0.736 J.

(c). The change in the gravitational potential energy of the system is 13.18 J

(d). The thermal energy of the system is 6.84 J.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus.

Answer is A.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus. Option a is correct.

A very rigid material—one that stretches or compresses only slightly under large forces—has a large value of Young's modulus. Young's modulus is a measure of the material's stiffness or rigidity. The larger the Young's modulus, the more stress is required to stretch the material to the same extent. For example, steel has a high Young's modulus, while rubber has a low Young's modulus.

Young’s Modulus and Material Rigidity

Young’s modulus, also known as the elastic modulus, is a measure of the stiffness of a material. It quantifies how much a material will deform (i.e., stretch or compress) under a given amount of force. A very rigid material is one that exhibits minimal deformation when subjected to large forces, and this characteristic is directly related to the value of Young’s modulus.

Young’s Modulus and Material Behavior

When a material is subjected to an external force, it deforms in response to that force. This deformation can take the form of stretching (tensile deformation) or compression (compressive deformation). Young’s modulus specifically measures the material’s response to tensile or compressive forces. A high value of Young’s modulus indicates that the material experiences minimal elongation or compression when subjected to these forces, signifying its rigidity.

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Answer:

[tex]EMF = 1684.67 Volts[/tex]

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

[tex]EMF = \frac{d\phi}{dt}[/tex]

now we have

[tex]\phi = BA[/tex]

here since magnetic field is constant so we have

[tex]EMF = A\frac{dB}{dt}[/tex]

now we have

[tex]A = (190 \times 10^3)(190 \times 10^3)[/tex]

[tex]A = 3.61 \times 10^{10} m^2[/tex]

now we have

[tex]EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})[/tex]

[tex]EMF = 1684.67 Volts[/tex]

Using Faraday's Law and given values,  the induced emf in a 190 km square loop subjected to a changing magnetic field. The calculated emf is approximately -1692 volts.

To find the emf induced in a square loop due to a changing magnetic field, we can use Faraday's Law of Electromagnetic Induction.

This law states that the magnitude of the induced emf around a closed loop is proportional to the rate of change of the magnetic flux through the loop.

Convert the rate of change of the magnetic field to Tesla per second (T/s):

2.8 μT/min = 2.8 × 10⁻⁶ T/min

Since there are 60 seconds in a minute:

r2.8 × 10⁻⁶ T/60 s ≈ 4.67 × 10⁻⁸ T/s

Calculate the area (A) of the square:

A = L × L = (190,000 m) × (190,000 m) = 3.61 × 10¹⁰ m²

Compute the induced emf (E) using Faraday's Law:

E = -A × dB/dt

E = - (3.61 × 10¹⁰ m²) × (4.67 × 10⁻⁸ T/s) ≈ -1692 V

Therefore, the emf induced in the square loop is approximately -1692 volts.

Answer:

Area of the rectangular field in kilometers is 0.09 [tex]km^2[/tex]

Explanation:

We know that 1 kilometers = 1000 meters

                    since we need to find the area in unit of kilometers

                    therefore converting length and width into kilometers

           1000 meters = 1 kilometers

             300 meters =[tex]\frac{300}{1000} = 0.3[/tex]

         Likewise  width  = 0.3 km

Area = length x width

         = 0.3km x 0.3 km

        = 0.09 [tex]km^2[/tex]

Answer:T=[tex]\frac{-26}{r}[/tex]+36

Explanation:

Given Temperature at r=1m is [tex]10^{\circ}C [/tex]

Temperature at r=2m is [tex]28^{\circ}C[/tex]

[tex]\frac{\mathrm{d^2} T}{\mathrm{d} r^2}+\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=0[/tex]

Let [tex]\frac{2}{r}\frac{\mathrm{d} T}{\mathrm{d} r}=S[/tex]

[tex]\frac{\mathrm{d^2} T}{\mathrm{d} r^2}=\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}[/tex]

therefore [tex]\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}+\frac{2}{r}[/tex]S=0

[tex]\frac{2}{r}\frac{\mathrm{d} S}{\mathrm{d} r}=-\frac{-2S}{r}[/tex]

solving

[tex]r^2[/tex] S=constant

substitute S value

[tex]\frac{\mathrm{d}T}{\mathrm{d}r}[/tex]=[tex]\frac{c}{r^2}[/tex]

Solving it we get

T=[tex]\frac{-c}{r}+c_2[/tex]

Now using given condition

10=[tex]\frac{-c}{1}+c_2[/tex]

28=[tex]\frac{-c}{2}+c_2[/tex]

[tex]c_2[/tex]=36,c=26

putting c values

T=[tex]\frac{-26}{r}[/tex]+36

The expression for the temperature T(r) for the two spheres one inside other is given as,

[tex]T=-\dfrac{26}{r}+36[/tex]

Differential equation is the equation in which there is one or more number of unknown variable exist to find the rate of change of one variable with respect to other.

The radius of the sphere is 1 m and the temperature of the sphere is 10 degree Celsius. The sphere lies inside a concentric sphere with radius 2 m and temperature 28°C.

The differential equation which spheres satisfies is given as,

[tex]\dfrac{d^2T}{dr^2}+\dfrac{2}{r}\dfrac{dT}{dr}=0[/tex]

Let the above equation is equation one,

Suppose S = dT/dr, then

[tex]\dfrac{dS}{dt}=\dfrac{2}{r}\dfrac{d^2T}{dr^2}[/tex]

Put the values in the equation one as,

[tex]\dfrac{2}{r}\dfrac{dS}{dt}+\dfrac{-2}{r}S=0\\\dfrac{2}{r}\dfrac{dS}{dt}=-\dfrac{-2}{r}S\\r^2S=C[/tex]

Here, (C) is the constant value. Now put the value of S as,

[tex]r^2\dfrac{dT}{dr}=C\\\dfrac{dT}{dr}=\dfrac{C}{r^2}\\T=-\dfrac{C}{r}+C_2[/tex]

Put the value of radius as 1 m,

[tex]10=-\dfrac{C}{1}+C_2\\10=-C+C_2[/tex]

Similarly, for the radius 2 meters and temperature 28 degree Celsius,

[tex]28=-\dfrac{C}{2}+C_2[/tex]

On solving above equation, we get,

[tex]C=26\\C_2=36[/tex]

Put the values of constant for the required expression as,

[tex]T=-\dfrac{26}{r}+36[/tex]

Thus, the expression for the temperature T(r) for the two spheres one inside other is given as,

[tex]T=-\dfrac{26}{r}+36[/tex]

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Answer:

a) The object must have constant velocity.

d) The object must have zero acceleration.

Explanation:

We can solve the problem by using Newton's second law, which states that the net force acting on an object is equal to the product between mass and acceleration:

[tex]F = ma[/tex]

where

F is the net force

m is the mass of the object

a is the acceleration

In this problem, the net force on the object is zero:

F = 0

This means that the acceleration of the object is also zero, according to the previous equation:

a = 0

So statement (d) is correct. Moreover, acceleration is defined as the rate of change of velocity:

[tex]a=\frac{\Delta v}{\Delta t}[/tex]

Which means that [tex]\Delta v=0[/tex], so the velocity is constant. Therefore, statement (a) is also correct. The other two statements are false because:

b)The object must be at rest. --> false, the object can be moving at constant velocity, different from zero

c)The object must be at the origin. --> false, since the object can be in motion

If the net force on an object is zero, the object might have constant velocity or zero acceleration. It's not necessary that the object be at rest or at the origin.

The subject of your question is related to the principles of Physics, specifically Newton's First Law of Motion. If the net force on an object is zero, it means that the object is in a state of equilibrium. From the given options:

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Answer:

A any air resistance

Explanation:

a body in freefall donot experience

Objects in free-fall motion, do not experience air resistance as it only moves under the effect of gravity. Therefore, option (A) is correct.

Free fall can be described as a motion of an object in which the force of gravity is the sole force acting upon it. An object moving upwards will not consider being falling. But if the body falls under the influence of gravity is said to be in free fall.

Free fall can be described as a type of motion in which no air resistance is considered and only gravity is considered. All bodies under free fall with the same rate of acceleration, regardless of their masses.

A body that falls through the air, has suffered some degree of air resistance. Air resistance can be described as the collisions of the surface of an object with gas molecules in the air. The factors that affect air resistance are the cross-sectional area and the speed of the body.

Learn more about free-fall motion, here:

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Answer:

The principal quantum number of the valence orbitals increases.

Explanation:

The atomic radius of main-group elements generally increases down a group because the principal quantum number of the valence orbitals increases.

The atomic radius of main-group elements increases down a group because the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus.

The atomic radius of main-group elements generally increases down a group because as you go down a column of the periodic table, the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus. This trend can be summarized as follows:

Given the same ramp angle and coefficient of friction, the lighter gurney (50kg) is more likely to slide down the ramp since it requires less friction force to stay stationary as compared to the heavier gurney (200kg).

The likelihood of the gurneys sliding down the slope depends upon the balance of forces on each. The force due to gravity on each gurney is mg sin θ, where m is mass, g is acceleration due to gravity, and θ is the angle of the ramp. The force of static friction on each gurney is μs N = μs mg cos θ, where μs is the coefficient of static friction, and N is the normal force, which equals mg cos θ.

In order for the gurneys not to slide, the friction force (μs mg cos θ) must be equal to or greater than the force due to gravity (mg sin θ). For a larger mass (like gurney 1 with the 200 kg dummy), the friction force is greater, so it is more likely to stay put. On the other hand, the smaller mass (gurney 2 with the 50 kg dummy) suffers less friction force due to its lesser weight. Therefore, as long as the angle of the ramp and the coefficient of friction are the same for both gurneys, gurney 2 (50 kg) is more likely to slide down the ramp.

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The current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

What is the current when t = 0.7 s?

Recall the relationship between current and charge:

Current (I) is the rate of change of charge (Q) over time. Mathematically, it's represented as:

I(t) = dQ(t)/dt

Differentiate the charge function:

Given the function for Q(t) = t³ - 2t² + 4t + 1, we can find the current I(t) by differentiating it with respect to time t:

I(t) = 3t² - 4t + 4

Calculate the current at t = 0.7 s:

Now, plug t = 0.7 into the expression for I(t):

I(0.7) = 3(0.7)² - 4(0.7) + 4

I(0.7) ≈ 2.67 C/s

Therefore, the current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]

Explanation:

Initial moment of inertia when arms and legs in is [tex]I_i=0.90 kg.m^{2}[/tex]

Final moment of inertia when her arms and on leg open outward, [tex]I_f=3.0 kg.m^{2}[/tex]

Initial angular speed [tex]w_i=5.2\frac{rev}{s}[/tex]

Let the final angular speed be [tex]w_f[/tex]

Since external torque on her is zero so we can apply conservation of angular momentum

[tex]\therefore L_f=L_i[/tex]

=>[tex]I_fw_f=I_iw_i[/tex]

=>[tex]w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}[/tex]

Thus her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]

Explanation:

It is known that density is the mass present in per unit volume.

Mathematically,         Density = [tex]\frac{mass}{volume}[/tex]

Since, it is given that [tex]d_{1}[/tex] is 1.18 [tex]kg/m^{3}[/tex], [tex]d_{2}[/tex] is 5.30 [tex]kg/m^{3}[/tex], and volume is 2 [tex]m^{3}[/tex].

Therefore, mass of air that has entered will be [tex]m_{2}[/tex] -  [tex]m_{1}[/tex] and it will be calculated as follows.

                       [tex]d_{2} - d_{1}[/tex] = [tex]\frac{m_{2} - m_{1}}{Volume}[/tex]

      [tex]m_{2} - m_{1}[/tex] = [tex](5.30 kg/m^{3} - 1.18 kg/m^{3}) \times 2 m^{3}[/tex]

                                            = 8.24 kg

Thus, we can conclude that mass of air that has entered the tank is 8.24 kg.

All points in a rigid body move with the same velocity and acceleration if the rigid body is subjected to pure translational motion.

When a rigid body is subjected to pure translational motion, all points in the body move with the same velocity and acceleration. This means each point in the body experiences the same linear velocity and acceleration at any given instant.

Answer:

The correct answer is kinetic energy. :) hope this helps!

Explanation:

Answer:

The length of the string is 0.266 meters.

Explanation:

It is given that,

Mass of the string, [tex]m=5.5\times 10^{-3}\ kg[/tex]

Tension in the string, T = 230 N

Frequency of wave, f = 160 Hz

Wavelength of the wave, [tex]\lambda=0.66\ m[/tex]

We need to find the length of the string. Let l is the length of the string. The speed of a transverse wave is given by :

[tex]v=\sqrt{\dfrac{T}{M}}[/tex]

M is the mass per unit length, M = m/l

[tex]v=\sqrt{\dfrac{lT}{m}}[/tex]

[tex]l=\dfrac{v^2m}{T}[/tex]

The velocity of a wave is, [tex]v=\nu\times \lambda[/tex]

[tex]l=\dfrac{(\nu\times \lambda)^2m}{T}[/tex]

[tex]l=\dfrac{(160\ Hz\times 0.66\ m)^2\times 5.5\times 10^{-3}\ kg}{230\ N}[/tex]

l = 0.266 meters

So, the length of the string is 0.266 meters. Hence, this is the required solution.                  

Answer:

L = 0.275 m

Explanation:

velocity of transverse wave in a stretched string is  given as

[tex]v =\sqrt \frac{T}{\mu}[/tex]

where T = tension = 230N

μ = linear density

[tex]μ = \frac[m}{L}[/tex]

where length L is in meters

Velocity = [tex]n\lambda[/tex]

so we have after equating both value of velocity

[tex]\sqrt \frac{T}{\mu} = n\lambda[/tex]

[tex]\frac{T}{\mu} =(n\lambda)^{2}[/tex]

μ  = [tex]\frac{T}{(n\lambda)^{2}}[/tex]

μ = [tex] \frac{230}{(160*0.66)^{2}}[/tex]

μ = 0.020 kg/m

but μ = [tex]\frac[m}{L}[/tex]

so length of string is

L = [tex]\frac{5.5*10^{-3}}{0.020}[/tex]

L = 0.275 m

Using the principle of conservation of momentum, the velocity of the 8 kg block, after the 4.3 kg block moves to the right with a speed of 6.7 m/s, is calculated to be -3.6 m/s. The negative sign denotes that the block is moving in the opposite direction to the 4.3 kg block.

The subject of this question is a part of physics known as mechanics, specifically conservation of momentum. The principle of conservation of momentum in a system where no external forces are acting states that the total momentum before an event must be equal to the total momentum after the event. Here, since the system begins with zero total momentum (both blocks initially at rest), it should end with zero total momentum.

In this scenario, after cord burns, the two blocks are free to move. The block with mass 4.3 kg moves to the right. According to the conservation of momentum, the other block will move in the opposite direction (to the left) in order to conserve the total momentum of the system.

We calculate the momentum of the system after the spring is released: Momentum = mass * velocity. For the 4.3 kg block, the momentum would be = 4.3 kg * 6.7 m/s = 28.81 kg*m/s. As the total momentum before the event was zero, the momentum of 8 kg block should be -28.81 kg*m/s (in opposite direction). The velocity of this block can now be calculated by dividing its momentum by its mass, i.e., -28.81 kg*m/s / 8 kg = -3.6 m/s (the negative sign indicates that the velocity is to the left).

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Using the formula for work (Work = Force x Distance), force and distance pairs are matched to their corresponding 'Work Performed' measurements. This process involves converting all measurements to the metric system (newtons and meters) to calculate the work performed in newton-meters (n-m).

The formula for work is Work = Force x Distance. Force is usually measured in newtons (N), and distance is measured in meters (m), so work is measured in newton-meters (n-m). One foot-pound (ft-lb) is equivalent to 1.35582 n-m and one pound is equivalent to 4.44822_newtons_ in the metric system. Therefore, the provided force and distance pairs can be matched to the correct 'Work Performed' answer as follows:

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Answer:

(c) no different than on a low-pressure day.

Explanation:

The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.

Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.  

Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.

Final answer:

The spring will stretch approximately 28 inches when the weight is increased to 44lb.

Explanation:

The length that a hanging spring stretches varies directly with the weight placed at the end of the spring. To find out how far the spring will stretch when the weight is increased to 44lb, we can set up a proportion based on the given information:

Proportion:

11lb / 7in. = 44lb / x

Using the cross multiplication method, we can solve for x:

11lb * x = 44lb * 7in.

x = 308in. / 11lb

x ≈ 28in.

Therefore, the spring will stretch approximately 28 inches when the weight is increased to 44lb.

Answer:

v = 46.55 m/s

Explanation:

It is given that,

A ball is thrown horizontally from the top of a building 0.10 km high, d = 0.1 km = 100 m

The ball strikes the ground at a point 65 m horizontally away from and below the point of release, h = 65 m

At maximum height, velocity of the ball is 0. So, using the equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g

[tex]100=0+\dfrac{1}{2}\times 9.8t^2[/tex]

[tex]t=4.51\ s[/tex]

Let [tex]v_x[/tex] is the horizontal velocity of the ball. It is calculated as :

[tex]v_x=\dfrac{65\ m}{4.51\ s}=14.41\ m/s[/tex]

Let [tex]v_y[/tex] is the final speed of the ball in y direction. It can be calculated as :

[tex]v_y^2+u_y^2=2as[/tex]

[tex]u_y=0[/tex]

[tex]v_y^2=2gd[/tex]

[tex]v_y^2=2\times 9.8\times 100[/tex]

[tex]v_y=44.27\ m/s[/tex]

Let v is the speed of the ball just before it strikes the ground. It is given by :

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

[tex]v=\sqrt{14.41^2+44.27^2}[/tex]

v = 46.55 m/s

So, the speed of the ball just before it strikes the ground is 46.55 m/s. Hence, this is the required solution.

The speed of the ball just before it strikes the ground is equal to 46.55 m/s.

Given the following data:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex].

To determine the speed of the ball just before it strikes the ground:

First of all, we would determine the time it took the ball to strike the ground by using the formula for maximum height.

[tex]H = \frac{1}{2} gt^2\\\\100 = \frac{1}{2} \times 9.8 \times t^2\\\\200 = 9.8t^2\\\\t^2 = \frac{200}{9.8} \\\\t^2=20.41\\\\t=\sqrt{20.41}[/tex]

Time, t = 4.52 seconds

Next, we would find the horizontal velocity:

[tex]Horizontal\;velocity = \frac{horizontal\;distance}{time} \\\\Horizontal\;velocity = \frac{65}{4.52}[/tex]

Horizontal velocity, V1 = 14.38 m/s

Also, we would find the velocity of the ball in the horizontal direction:

[tex]V_2^2 = U^2 + 2aS\\\\V_2^2 = 0^2 + 2(9.8)(100)\\\\V_2^2 = 1960\\\\V_2 = \sqrt{1960} \\\\V_2 = 44.27 \;m/s[/tex]

Now, we would calculate the speed of the ball just before it strikes the ground by finding the resultant speed:

[tex]V = \sqrt{V_1^2 + V_2^2} \\\\V = \sqrt{14.38^2 + 44.27^2}\\\\V = \sqrt{206.7844‬ + 1959.8329‬}\\\\V =\sqrt{2166.6173}[/tex]

Speed, V = 46.55 m/s

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