A rope attached to a load of 175 kg bricks Ilifts the bricks with a steady acceleration of 0.12.m/s^2 straight up. What is the tension in the rope? (a)2028N (b)1645 N (c) 1894 N (d) 1976 N (e) 1736 N (f) 1792 N

Answer:

3.3 m/s

Explanation:

As the object rises above the lowest point, some of the kinetic energy is converted to potential energy.  From the diagram, we can see that at angle θ, the object rises to height h:

h = L - L cos θ

Conservation of energy:

KE₀ = KE + PE

KE₀ = 1/2 mv² + mgh

Substituting:

KE₀ = 1/2 mv² + mg(L - L cos θ)

Given KE₀ = 10.0 J, m = 0.80 kg, g = 9.8 m/s², L = 2.0 m, and θ = 50.0°:

10.0 = 1/2 (0.80) v² + (0.80) (9.8) (2.0 - 2.0 cos 50.0)

v = 3.32 m/s

Rounding to 2 sig-figs, the speed of the object is 3.3 m/s.

Answer:

The distance is 8.25 m.

Explanation:

Given  that,

Speed = 2 m/s

Acceleration = 0.5 m/s^2

Time = 3 sec

We need to calculate the distance

Using equation of motion

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

Where, u = initial velocity

a = acceleration

t = time

Put the value in the equation

[tex]s=2\times3+\dfrac{1}{2}\times0.5\times3^2[/tex]

[tex]s=8.25\ m[/tex]

Hence, The distance is 8.25 m.

Julie's distance at the end of the 3 seconds is 8.25 m.

To calculate the distance traveled by Julie at the end ]of 3 seconds, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these given values into equation 1

Hence, Julie's distance at the end of the 3 seconds is 8.25 m.
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Answer:

Maximum height reached by the bullet = 4.01 m

Explanation:

Horizontal displacement = 92 m

Time taken = 6.4 s

Horizontal velocity

           [tex]=\frac{92}{6.4}=14.375m/s[/tex]

We have angle of projection = 32°

Horizontal velocity = u cos 32 = 14.375

                      u = 16.95 m/s

Vertical velocity = u sin θ = 16.95 x sin 32 = 8.98 m/s

Time of flight till it reaches maximum height = 0.5 x 6.4 = 3.2s

Now we have vertical motion of bullet

        S = ut + 0.5 at²

       Vertical velocity = u = 16.95 m/s

       a = -9.81 m/s²

        t = 3.2s

Substituting

        S = 16.95 x 3.2 - 0.5 x 9.81 x 3.2² = 4.01 m

Maximum height reached by the bullet = 4.01 m

The magnitude of linear momentum can be calculated using the equation p=mv. The momentum for the different objects given are: (a) Proton: 8.517 * 10^-21 kg*m/s, (b) Bullet: 7.2 kg*m/s, (c) Sprinter: 761.25 kg*m/s, and (d) Earth: 1.78324 * 10^29 kg*m/s.

The magnitude of the linear momentum of an object is calculated using the equation p=mv, where 'p' is the momentum, 'm' is the mass of the object, and 'v' is its velocity.

(a) Using the given values, m = 1.67 * 10^-27 kg and v = 5.10 * 10^6 m/s, the momentum of the proton is p = m*v = (1.67 * 10^-27 kg)*(5.10 * 10^6 m/s) = 8.517 * 10^-21 kg*m/s.

(b) The bullet's mass is 15.0 g which is 0.015 kg. Its velocity is 480 m/s. Thus, its momentum is p = m*v = (0.015kg)*(480m/s) = 7.2 kg*m/s.

(c) The sprinter's mass is 72.5 kg and velocity is 10.5 m/s. So, his momentum is p = m*v = (72.5kg)*(10.5m/s) = 761.25 kg*m/s.

(d) The Earth's mass is 5.98 * 10^24 kg and its orbital speed is 2.98 * 10^4 m/s. Hence, its momentum is p = m*v = (5.98 * 10^24 kg)*(2.98 * 10^4 m/s) = 1.78324 * 10^29 kg*m/s.

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The magnitude of the linear momentum for a proton, a bullet, a sprinter, and the Earth is approximately 8.52 * 10^-21 kg*m/s, 7.2 kg*m/s, 760.5 kg*m/s, and 1.78 * 10^29 kg*m/s, respectively.

To calculate the magnitude of the linear momentum, we need to multiply the mass of the object by its velocity, which follows the formula: momentum = mass * velocity.

For the given cases:

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Answer:

[tex]I = 94.33 kg m^2[/tex]

Explanation:

Let say the rod is slightly pulled away from its equilibrium position

So here net torque on the rod due to its weight is given as

[tex]\tau = mg dsin\theta[/tex]

since rod is pivoted at distance of 1.4 m from centre of gravity

so its moment of inertia about pivot point is given as

[tex]Inertia = I[/tex]

now we have

[tex]I \alpha = mg d sin\theta[/tex]

now for small angular displacement we will have

[tex]\alpha = \frac{mgd}{I}\theta[/tex]

so angular frequency of SHM is given as

[tex]\omega = \sqrt{\frac{mgd}{I}}[/tex]

now we will have

[tex]\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}[/tex]

[tex]I = 94.33 kg m^2[/tex]

Final answer:

The moment of inertia from the pivot is approximately [tex]72.4 kg\(\cdot\)m2.[/tex]

Explanation:

To generate an accurate answer for the moment of inertia of the stick about the pivot when it is used as a physical pendulum, we can use the formula for the period of a physical pendulum, which is:

[tex]T = 2\(\pi\)\(\sqrt{I/(mgd)}\),[/tex]

where T is the period, I is the moment of inertia, m is the mass of the rod, g is the acceleration due to gravity (approximately 9.8 m/s2), and d is the distance from the pivot to the center of mass. Given that T = 9 seconds, m = 4.8 kg, and d = 1.4 m, we can rearrange the formula to solve for I:

[tex]I = T2mgd/(4\(\pi\)2).[/tex]

Plugging the given values into this equation:

[tex]I = 92 * 4.8 kg * 9.8 m/s^2 * 1.4 m / (4 * \(\pi\)2).[/tex]

Calculating this yields the moment of inertia I:

[tex]I \(\approx\) 72.4 kg\(\cdot\)m2.[/tex]

Answer:

A. The starting height of the ball

Explanation:

When we talk about controlled variables, we refer to the variable that should be kept the same throughout the experiment. The reason why we do this, is to limit anything else that is not being tested, that may affect the results of the experiment.

In the scenario given, the experiment is to see the relationship between the initial height of a basketball and the height of its rebound bounce.

So you the starting height of the ball should vary, meaning it is NOT controlled.

Answer:

Starting height of the ball

Explanation:

Answer:

I think it will be half of the initial charge

Explanation:

Because we know, the resulting charge will be q1+q2/2, since one is neutral so the charge will be half q/2

Answer:

d = 15 mm

Explanation:

Force of repulsion between two current carrying wire is given by

[tex]F = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

now this force of repulsion is counterbalanced by the weight of the wire

so we have

[tex]mg = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

now we have

[tex]d = \frac{\mu_0 i_1 i_2 L}{2\pi mg}[/tex]

so here we can say that

[tex]d = \frac{\mu_0 i_1 i_2}{2\pi (m/L)g}[/tex]

now plug in all values in it

[tex]d = \frac{4\pi \times 10^{-7} (72.8)(72.7)}{2\pi (0.071)}[/tex]

[tex]d = 0.015 m[/tex]

d = 15 mm

(a) [tex]4.3\cdot 10^9 J[/tex]

The kinetic energy of an object is given by:

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

where

m is the mass of the object

v is its speed

For the ship in the problem, we have

[tex]m=6.40\cdot 10^7 kg[/tex] is the mass

[tex]v=11.6 m/s[/tex] is the speed

So its kinetic energy is

[tex]K=\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=4.3\cdot 10^9 J[/tex]

(b) [tex]-4.3\cdot 10^9 J[/tex]

According to the work-kinetic energy theorem, the work done on an object is equal to the change in kinetic energy of the object:

[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]

where

W is the work done

m is the mass of the object

v is the final speed of the object

u is the initial speed of the object

Here we want to find the work done to stop the ship, so the final speed of the ship is v=0, while the initial speed is u=11.6 m/s. So the work done will be

[tex]W= 0 -\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=-4.3\cdot 10^9 J[/tex]

(c) [tex]1.3\cdot 10^6 N[/tex]

The work done on an object can be also written as follows

[tex]W=Fd[/tex]

where

F is the magnitude of the force

d is the displacement of the object

Here we know:

[tex]W=-4.3\cdot 10^9 J[/tex] is the work done

d = 3.20 km = 3200 m is the displacement of the ship

So we can solve the formula to find F, the force exerted on the ship to stop it:

[tex]F=\frac{W}{d}=\frac{-4.3\cdot 10^9 J}{3200 m}=-1.3\cdot 10^6 N[/tex]

and the negative sign simply means that the force is opposite to the displacement of the ship (in fact, the force acts against the motion of the ship).

The ship's kinetic energy is 5.77 × 10^9 J. The work required to stop it is -5.77 × 10^9 J. The magnitude of the force required to stop it is 1.80 × 10^3 N.

(a) The kinetic energy of the ship can be calculated using the formula:

Ek = (1/2)mv2

where m is the mass of the ship and v is its velocity. Plugging in the given values, we get a kinetic energy of 5.77 × 109 J.

(b) To stop the ship, work needs to be done to bring it to a complete stop. The work done is equal to the change in kinetic energy of the ship. Initially, the ship had a kinetic energy of 5.77 × 109 J. When it comes to a stop, its kinetic energy becomes zero. Therefore, the work required to stop the ship is -5.77 × 109 J (negative since work is done on the ship).

(c) The magnitude of the constant force required to stop the ship can be calculated using the work-energy theorem:

Work = Force × Distance

Given that the displacement of the ship is 3.20 km, or 3.20 × 106 m, and the work done on the ship is -5.77 × 109 J, we can solve for the force:

Force = Work / Distance = -5.77 × 109 J / (3.20 × 106 m) = -1.80 × 103 N

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Answer:

Magnitude of force, F = 11 Newtons

Explanation:

Charge 1, [tex]q_1=5.3\ \mu C=5.3\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=11.2\ \mu C=11.2\times 10^{-6}\ C[/tex]

The separation between the charges is, r = 0.22 m

We have to find the magnitude of the force between two point charges. It can be calculated using the formula as :

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

[tex]F=9\times 10^9\times \dfrac{5.3\times 10^{-6}\ C\times 11.2\times 10^{-6}\ C}{(0.22\ m)^2}[/tex]

F = 11.03 N

or

F = 11 Newtons

Hence, this is the required solution.

Answer:

Final velocity, v = 25.3 m/s

Explanation:

Initial velocity of a locomotive, u = 19 m/s

Acceleration of the locomotive, a = 0.8 m/s²

Length of station, d = 175 m

We need to find its final velocity (v) when the nose leaves the station. It can be calculated using the third law of motion :

[tex]v^2-u^2=2ad[/tex]

[tex]v^2=2ad+u^2[/tex]

[tex]v^2=2\times 0.8\ m/s^2\times 175\ m+(19\ m/s)^2[/tex]

[tex]v^2=(641)\ m^2[/tex]

v = 25.31 m/s

v = 25.3 m/s

When the nose leaves the station, it will move with a velocity of 25.3 m/s. Hence, this is the required solution.

Answer:

Induced emf of the wire is 6.36 Volts.

Explanation:

It is given that,

Length of the wire, l = 75 cm = 0.75 m

Magnetic field, B = 0.53 T

Velocity, v = 16 m/s

The wire is moving straight up in the magnetic field. So, an emf is induced in the wire. It is given by :

[tex]\epsilon=Blv[/tex]

[tex]\epsilon=0.53\ T\times 0.75\ m\times 16\ m/s[/tex]

[tex]\epsilon=6.36\ V[/tex]

So, the induced emf of the wire is 6.36 V. Hence, the correct option is (b) "6.36 V".

Answer:

6.36

Explanation:

If a 75 cm straight wire moves straight up through a 0.53 T magnetic field with a velocity of 16 m/s, the induced emf in the wire is 6.36.

Start with 2 arbitrary vectors, [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex]. (pic 1)

Vectors are determined by their lengths and direction. This means that translating the vector (i.e. sliding it left/right and up/down in the plane) doesn't fundamentally change that vector. To this end, we could just as easily represent [tex]\vec v_2[/tex] as if it had originated from the tip of [tex]\vec v_1[/tex]. This "new" [tex]\vec v_2[/tex] and the "old" [tex]\vec v_2[/tex] are the same vector. (pic 2)

If we connect the origin of [tex]\vec v_1[/tex] with the tip of "new" [tex]\vec v_2[/tex], we get a new vector, and this we define as the vector sum [tex]\vec v_1+\vec v_2[/tex]. (pic 3)

We can do this other way, by first traslating [tex]\vec v_1[/tex] to the tip of [tex]\vec v_2[/tex], then connecting the origin of [tex]\vec v_2[/tex] with the tip of "new" [tex]\vec v_1[/tex]. This demonstrates that vector addition is commutative (order of the vectors being added doesn't matter - you always end up at the same terminus). The "parallelogram method" refers to how a parallelogram is traced out. (pic 4)

Multiplying a vector by -1 reverses its direction. (pic 5)

Adding [tex]\vec v_1[/tex] and [tex]-\vec v_2[/tex] works the same way as standard vector addition, giving us the new vector [tex]\vec v_1-\vec v_2[/tex]. (pic 6)

We can do the same in the reverse order, but now we get a different vector, [tex]\vec v_2-\vec v_1[/tex]. (pic 7)

These vectors have the same length but point in opposite directions. (pic 8)

But notice that we can translate the vectors [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] so that we get a vector that either starts at the tip of [tex]\vec v_2[/tex] and ends at the tip of [tex]\vec v_1[/tex] (pic 9), or starts at the tip of [tex]\vec v_1[/tex] and ends at the tip of [tex]\vec v_2[/tex] (pic 10). The "triangle method" refers to the triangles that are traced out by either vector sum [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] together with [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex].

Answer:

The number of protons 6.19 more than electron.

Explanation:

Given that,

Charge [tex]Q=+9.9\times10^{-6}\ C[/tex]

We know that,

formula of charge

[tex]Q = nc[/tex]

Where,

Q = total charge

n = number of protons

e = charge of electron

Put the value into the formula

[tex]n = \dfrac{Q}{e}[/tex]

[tex]n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}[/tex]

[tex]n=6.1875\times10^{13}\ C[/tex]

According to statement of question

Divide the answer by [tex]10^{13}[/tex]

[tex]n=\dfrac{6.1875\times10^{13}}{10^{13}}[/tex]

[tex]n=6.19\ C[/tex]

Hence, The number of protons 6.19 more than electron.

The shell has horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] at time [tex]t[/tex] according to

[tex]x=\left(310\dfrac{\rm m}{\rm s}\right)\cos50.0^\circ t[/tex]

[tex]y=\left(310\dfrac{\rm m}{\rm s}\right)\sin50.0^\circ t-\dfrac g2t^2[/tex]

where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.

The shell explodes after 39.0 s, at which point its coordinates are

[tex]x=7770\,\mathrm m[/tex]

[tex]y=1810\,\mathrm m[/tex]

Answer:

Maximum height, h = 1.74 meters

Explanation:

It is given that,

A potato is shot out of the cylinder. It is a case of projectile motion. The potato makes an angle of 17 degrees above the horizontal.

Initial speed with which the potato is shot out, u = 20 m/s

We have to find the maximum height of the potato. The maximum height of a projectile (h) is given by the following formula as :

[tex]h=\dfrac{u^2sin^2\theta}{2g}[/tex]

Where

[tex]\theta[/tex] = angle between the projectile and the surface

g = acceleration due to gravity

[tex]h=\dfrac{(20\ m/s)^2sin^2(17)}{2\times 9.8\ m/s^2}[/tex]

h = 1.74 m

or h = 1.74 meters

Hence, this is the required solution.

Answer:

15.Radiowave

16.laser is device that generates an intense beam of other electromagnetic radiation by emission of photons from excited atoms.

17.this is a laboratory instrument commonly used to display and analyse the waveformof electronic signals.

19. this is a space entirely devoid of matter.

To achieve a temperature independent resistance of 3.80 kΩ, you need to use a carbon resistor and Nichrome wire-wound resistor that counterbalance each other. This is possible because carbon and Nichrome have opposite temperature coefficients of resistance.

In order for the resistance to remain constant with temperature, the carbon resistor and the Nichrome wire-wound resistor must counterbalance each other. Meaning, when one's resistance increases with temperature, the other's resistance decreases, keeping the total resistance the same. Given that carbon and Nichrome have opposite temperature coefficients of resistance, they can accomplish this task.

Generally, the resistance R of a resistor is given by the formula R = R0(1 + α(T-T0)), where α is the temperature coefficient, T is the temperature and R0 is the resistance at reference temperature T0. As the temperature increases, a positive α will increase the resistance while a negative α will decrease it.

To make the combined resistance temperature independent, the sum of the change in resistance of the carbon resistor and the Nichrome resistor should be zero. Therefore, you would set up the equation where the increase of the carbon resistance equals the decrease of the Nichrome resistance. Solving this equation will give you the exact values required for the resistances of carbon and Nichrome at 0ºC in order to have a total resistance of 3.80 kΩ.

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Answer:

Orbital period, T = 2.02 hours

Explanation:

It is given that, an artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s². We have to find the orbital period (T) of the satellite.

Firstly, calculating the distance between Earth and satellite. The acceleration due to gravity is given by :

[tex]a=\dfrac{GM}{r^2}[/tex]

G = universal gravitational constant

M = mass of earth

[tex]r=\sqrt{\dfrac{GM}{a}}[/tex]

[tex]r=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{6.03\ m/s^2}}[/tex]

r = 8126273.3 m..........(1)

Now, according to Kepler's third law :

[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]

Putting the value of r from equation (1) in above equation as :

[tex]T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 5.97\times 10^{24}}\times (8126273.3)^3[/tex]

[tex]T^2=53202721.01\ s[/tex]

T = 7294.01 seconds

Since, 1 hour = 3600 seconds

Converting seconds to hour we get :

So, T = 2.02 hour  

So, the orbital period of the satellite is 2.02 hours.

Answer:

3.8 x 10⁴ Ω

Explanation:

C = Capacitance = 165 μF

R = resistance of the filament = ?

t = time taken to charge the capacitor = 10 s

Q₀ = maximum charge stored by capacitor

Q = charge stored by capacitor at time "t" = 0.80 Q₀

T = Time constant

Charge stored by capacitor at any time is given as

[tex]Q = Q_{o}(1 - e^{\frac{-t}{T}})[/tex]

[tex]0.80 Q_{o} = Q_{o}(1 - e^{\frac{-10}{T}})[/tex]

T = 6.21 s

Time constant is given as

T = RC

6.21 = R (165 x 10⁻⁶)

R = 3.8 x 10⁴ Ω

The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω

It is given that:-

Capacitance  C= 165 μF

Resistance of the filament = R=?

Time taken to charge the capacitor = t = 10 s

Now,

Q₀ = maximum charge which can be stored by the capacitor

Since the capacitor to 80% of its maximum charge

Then,

Q = charge stored by capacitor at time "t" = 0.80 Q₀

T = Time constant

The charge stored by the capacitor at any time is given as

[tex]Q=Q_{0} (1-e^{\dfrac{-t}{T} } )[/tex]

[tex]0.80Q_{0} =Q_{0} (1-e^{\dfrac{-10}{T} } )[/tex]

[tex]T=6.21 sec[/tex]  

Now the Time constant is given as

[tex]T=R\times C[/tex]

[tex]6.21= R\times (165\times 10^{-6} )[/tex]

[tex]R=3.8\times 10^{4}[/tex]Ω

The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω

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(a) The pitcher must throw the ball at 27.7 m/s

The momentum of an object is given by:

[tex]p=mv[/tex]

where

m is the mass of the object

v is the object's velocity

Let's calculate the momentum of the bullet, which has a mass of

m = 2.70 g = 0.0027 kg

and a velocity of

[tex]v=1.50\cdot 10^3 m/s[/tex]

Its momentum is:

[tex]p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s[/tex]

The pitcher must throw the baseball with this same momentum. The mass of the ball is

m = 0.146 kg

So the velocity of the ball must be

[tex]v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s[/tex]

So, the pitcher must throw the ball at 27.7 m/s.

(b) a. The bullet has greater kinetic energy

The kinetic energy of an object is given by

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

where m is the mass of the object and v is its speed.

For the bullet, we have:

[tex]K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J[/tex]

For the ball:

[tex]K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J[/tex]

So, the bullet has greater kinetic energy.

Final answer:

The baseball's speed must be 27.74 m/s to match the momentum of the bullet, and the bullet has greater kinetic energy than the baseball.

Explanation:

Calculating Baseball's Speed and Comparing Kinetic Energies

To validate the pitcher's claim that a 0.146-kg baseball can have the same momentum as a 2.70-g bullet traveling at 1.50 x 10³ m/s, we must use the formula for momentum (p = mv), which gives us:

Momentum of bullet = (2.70 g) x (1.50 x 10³ m/s) = (0.0027 kg) x (1500 m/s) = 4.05 kg m/s.

Therefore, the baseball's speed v needed to have the same momentum is calculated by rearranging the formula to v = p/m, which gives us:

Baseball's speed v = 4.05 kg m/s / 0.146 kg = 27.74 m/s.

To determine which has the greater kinetic energy, we use the kinetic energy formula KE = (1/2)mv². Calculating the kinetic energy of both the bullet and the baseball:

KE of bullet = (1/2)(0.0027 kg)(1500 m/s)² = 3037.5 J.

KE of baseball = (1/2)(0.146 kg)(27.74 m/s)² = 55.90 J.

Comparing the results shows that the bullet has greater kinetic energy than the baseball.

Answer:

Centripetal acceleration of the car is (4.96 g) m/s²

Explanation:

It is given that,

Radius of circle, r = 95 m

Speed of the car, v = 68 m/s

We need to find the centripetal acceleration. It is given by :

[tex]a_c=\dfrac{v^2}{r}[/tex]

So, [tex]a_c=\dfrac{(68\ m/s)^2}{95\ m}[/tex]

[tex]a_c=48.67\ m/s^2[/tex]

Since, g = 9.8 m/s²

So,

[tex]a_c=(4.96\ g)\ m/s^2[/tex]

So, the magnitude of the centripetal acceleration is (4.96 g) m/s². Hence, this is the required solution.

Answer:

24.14 ⁰C

Explanation:

T₀ = initial temperature = 3 °C

R₀ = initial resistance of thermometer at initial temperature = 217.62 Ω

R = Final resistance of thermometer at final temperature = 215.32 Ω

T = final temperature = ?

α = temperature coefficient of resistivity for carbon = - 5.00 x 10⁻⁴ C⁻¹

Final resistance of thermometer at final temperature is given as

R = R₀ (1 + α (T - T₀ ))

Inserting the values

215.32 = 217.62 (1 + (- 5.00 x 10⁻⁴) (T - 3))

T = 24.14 ⁰C

Explanation:

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / F

F = 50 / cos 28°

F = 56.6 lbs

56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change in momentum.

As given in the problem if a ramp leading to the entrance of a building is inclined upward at an angle of 7 degrees. A suitcase is to be pulled up the ramp by a handle that makes an angle of 35 degrees horizontal.

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / Force

Force  = 50 / cos 28°

Force  = 56.6 lbs

Thus, the 56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs

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Answer:

(a) 6.8 x 10^5 Nm^2/C

(b) 1.47 x 10^5 Nm^2/C

(c) 5.3 x 10^5 Nm^2/C

Explanation:

According to the Gauss's theorem

Electric flux = Charge enclosed / ∈0

(a) Charge enclosed = 6 x 10^-6 C

So, Electric flux = (6 x 10^-6) / (8.854 x 10^-12) = 6.8 x 10^5 Nm^2/C

(b) Charge enclosed = -1.3 x 10^-6 C

So, Electric flux = (1.3 x 10^-6) / (8.854 x 10^-12) = 1.47 x 10^5 Nm^2/C

(c) Charge enclosed = 6 x 10^-6 + (-1.3 x 10^-6) = 4.7 x 10^-6 C

So, Electric flux = (4.7 x 10^-6) / (8.854 x 10^-12) = 5.3 x 10^5 Nm^2/C

The electric flux through a spherical surface due to enclosed charges can be computed using Gauss's Law. The flux for a +6.60 x 10^-6 C charge is outward-directed, for a -1.30 x 10^-6 C charge it is inward-directed, and with both charges, the net flux is the sum of the individual fluxes.

The student is asking about the concept of electric flux through a spherical surface that surrounds a collection of charges, which falls under the subject of Physics (specifically electromagnetism), and it is a high school- or introductory college-level question. According to Gauss's Law, the electric flux through a closed surface is directly proportional to the enclosed electric charge. This can be calculated using the formula Φ = q/ε0, where Φ is the electric flux, q is the electric charge, and ε0 is the permittivity of free space (approximately 8.85 x 10^-12 C2/N⋅m2).

For part (a), a spherical surface surrounding a single +6.60 × 10-6 C charge would result in an outward-directed electric flux Φ = +6.60 × 10^-6 C / 8.85 × 10^-12 C2/N⋅m2.

For part (b), a spherical surface surrounding a single -1.30 × 10-6 C charge would have an inward-directed electric flux Φ = -1.30 × 10^-6 C / 8.85 × 10^-12 C2/N⋅m2.

For part (c), when both charges are enclosed, their net flux through the surface is the sum of the individual fluxes. Therefore the net electric flux is Φ = (+6.60 × 10^-6 C - 1.30 × 10^-6 C) / 8.85 × 10^-12 C2/N⋅m2, which simplifies to the sum of the charges divided by the permittivity of free space.

Answer:

[tex]E=3.3\times 10^4N/C[/tex]

Option D is the correct answer.

Explanation:

Electric field, E is the ratio of potential difference and distance between them.

Potential difference, V = 227 V

Distance between plates = 6.8 mm = 0.0068 m

Substituting,

         [tex]E=\frac{V}{d}=\frac{227}{0.0068}=3.3\times 10^4N/C[/tex]

Option D is the correct answer.

Answer:

27.22 K

Explanation:

T₁ = initial temperature in fahrenheit = - 4.00 ⁰F

T₂ = final temperature in fahrenheit = 45.0 ⁰F

To convert the temperature from fahrenheit to kelvin, we can use the relation

[tex]K = \frac{F - 32}{1.8} + 273.15[/tex]

where F = Temperature in fahrenheit  and K = temperature in kelvin

T'₁ =  initial temperature in kelvin = (- 4.00 - 32)/1.8 + 273.15 = 253.15 K

T'₂ =  final temperature in kelvin = (45.0 - 32)/1.8 + 273.15 = 280.37 K

ΔT = Change in temperature

Change in temperature on kelvin scale is given as

ΔT = T'₂ - T'₁

ΔT = 280.37 - 253.15

ΔT = 27.22 K

To find the temperature change on the Kelvin scale, convert the given temperatures from Fahrenheit to Kelvin and subtract them.

The temperature change on the Kelvin scale can be determined by converting the given temperatures from Fahrenheit to Kelvin and then finding the difference between them.

First, convert -4.00°F to Kelvin:
273.15 K + (-4.00°F + 459.67 °F) × (5/9)

Next, convert 45.0°F to Kelvin:
273.15 K + (45.0°F + 459.67 °F) × (5/9)

Finally, subtract the two Kelvin temperatures to find the temperature change.

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Answer:

Force exerted, F = 2.64 × 10⁷ Newton

Explanation:

It is given that,

Mass of the artillery shell, m = 1100 kg

It is accelerated at, [tex]a=2.4\times 10^4\ m/s^2[/tex]

We need to find the magnitude of force exerted on the ship by the artillery shell. It can be determined using Newton's second law of motion :

F = ma

[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]

F = 26400000 Newton

or

F = 2.64 × 10⁷ Newton

So, the force exerted on the ship by the artillery shell is 2.64 × 10⁷ Newton.

Answer: The force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex]  and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]

Explanation:

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

[tex]F=ma[/tex]

where,

F = force exerted on the artillery shell

m = mass of the artillery shell = 1100 kg

a = acceleration of the artillery shell = [tex]2.40\times 10^4m/s^2[/tex]

Putting values in above equation, we get:

[tex]F=1100kg\times 2.40\times 10^4m/s^2\\\\F=2.64\times 10^6N[/tex]

Now, according to Newton's third law, every action has an equal and opposite reaction.

So, the force exerted on the artillery shell will be equal to the force exerted on the ship by artillery shell acting in opposite direction.

Hence, the force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex]  and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]

Answer:

1.85 m

Explanation:

The horizontal velocity of the ball is

[tex]v_x = v cos \theta = (6.6 m/s) cos 39.8^{\circ}=5.1 m/s[/tex]

The horizontal distance travelled is

d = 2.7 m

And since the motion along the horizontal direction is a uniform motion, the time taken is

[tex]t= \frac{d}{v_x}=\frac{2.7 m}{5.1 m/s}=0.53 s[/tex]

The vertical position of the ball is given by

[tex]y= h + u_y t - \frac{1}{2}gt^2[/tex]

where

h = 1 m is the initial heigth

[tex]u_y = v sin \theta = (6.6 m/s) sin 39.8^{\circ}=4.2 m/s[/tex] is the initial vertical velocity

g = 9.82 m/s^2 is the acceleration due to gravity

Substituting t = 0.53 s, we find the height of the ball at this time:

[tex]y=1 m + (4.2 m/s)(0.53 s) - \frac{1}{2}(9.82 m/s^2)(0.53 s)^2=1.85 m[/tex]

Answer:

To convert kilometers per hour to meters per second, perform dimensional analysis. Remember that:

1 km = 1000 m

1 hr = 3600 seconds

Using these conversion, perform dimensional analysis:  

16.2 km/ hr (1000m/1km) (1hr/60 sec) = 4.5 m/s

The analysis basically just uses the conversion factors and canceling of units. The final answer is 4.5 m/s.  

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Correction: That should be *(1 hr/3600 sec). The answer is still 4.5 m/s.

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Hope this helps, i did the test and this answer was right, oh and brainliest, Good luck.

Answer:

The work is required to stretch it from 8 m to 16 m is 192 N-m

Explanation:

Given that,

Natural length = 8 m

Force F = 12 N

After stretched,

length = 10 m

We need to calculate the elongation

[tex]x = 10-8=2\ m[/tex]

Using hook's law

The restoring force is directly proportional to the displacement.

[tex]F\propto (-x)[/tex]

[tex]F = -kx[/tex]

Where, k = spring constant

Negative sign shows the displacement in opposite direction

Now, The value of k is

[tex]k = \dfrac{F}{x}[/tex]

[tex]k = \dfrac{12}{2}[/tex]

[tex]k = 6[/tex]

When stretch the string from 8 m to 16 m.

Then the elongation is

[tex]x=16-8=8\ m[/tex]

Now, The work is required to stretch it from 8 m to 16 m

[tex]W = \dfrac{1}{2}kx^2[/tex]

Where, k = spring constant

x = elongation

[tex]W=\dfrac{1}{2}\times6\times8\times8[/tex]

[tex]W=192\ N-m[/tex]

Hence, The work is required to stretch it from 8 m to 16 m is 192 N-m