Heavy water usually refers to water that : (A) has been frozen, and so is more dense (B) has had its hydrogen removed (C) is radioactive (D) contains deuterium

[tex]x(t)=7.00\,\mathrm m-\left(9.00\dfrac{\rm m}{\rm s}\right)t+\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

a. The particle has velocity at time [tex]t[/tex],

[tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=-9.00\dfrac{\rm m}{\rm s}+\left(6\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

so that after [tex]t=1\,\mathrm s[/tex] it will have velocity [tex]\boxed{-3.00\dfrac{\rm m}{\rm s}}[/tex].

b. The sign of the velocity is negative, so it's moving in the negative [tex]x[/tex] direction.

c. Its speed is 3.00 m/s.

d. The particle's velocity changes according to

[tex]\dfrac{\mathrm d^2x(t)}{\mathrm dt^2}=6\dfrac{\rm m}{\mathrm s^2}[/tex]

which is positive and indicates the velocity/speed of the particle is increasing.

e. Yes. The velocity is increasing at a constant rate. Solving for [tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=0[/tex] is trivial; this happens when [tex]\boxed{t=1.50\,\mathrm s}[/tex].

f. No, the velocity is positive for all [tex]t[/tex] beyond 1.50 s.

Answer:

141

Explanation:

The atomic number (Z) corresponds to the number of protons:

Z = p

while the mass number (A) corresponds to the number of protons+neutrons:

A = p + n

So the number of neutrons in a nucleus is equal to the difference between mass number and atomic number:

n = A - Z

For the initial nucleus of Uranium, Z = 92 and A = 235, so the initial number of neutrons is

n = 235 - 92 = 143

An alpha particle carries 2 protons and 2 neutrons: so, when the isotope of Uranium emits an alpha particle, it loses 2 neutrons. Therefore, the number of neutrons after the decay will be

n = 143 - 2 = 141

Final answer:

The type of friction acting between the tires and the road when a car's tires roll freely without any slippage is static friction, as the tire's contact point with the road is momentarily at rest.

Explanation:

When a car's tires roll freely without any slippage, the type of friction acting between the tires and the road is static friction. This is because the bottom of the tire is at rest with respect to the ground for a moment in time, ensuring there's no relative movement between the contact surfaces. Static friction is what allows the car to move forward as it prevents the tires from slipping on the surface of the road. Once slipping occurs, for instance, if the tires are spinning without moving the car forward, it becomes kinetic friction. However, in a scenario where rolling without slipping occurs, it's due to the presence of static friction, which is necessary for proper motion and control of the vehicle.

Answer:

D. None of the above

Explanation:

There are only two forces acting on a pendulum:

- The force of gravity (downward)

- The tension in the string

We can consider the axis along the direction of the string: here we have the tension T, acting towards the pivot, and the component of the weight along this direction, acting away from the pivot. Their resultant must be equal to the centripetal force, so we can write:

[tex]T-mg cos \theta = m\frac{v^2}{r}\\T=m\frac{v^2}{r}+mg cos \theta[/tex]

where

T is the tension in the string

[tex]\theta[/tex] is the angle between the tension and the vertical

m is the mass

g is the acceleration of gravity

v is the speed of the pendulum

r is the length of the string

From the formula we see that the value of the tension, T, depends only on the value of v (the speed) and [tex]\theta[/tex], the angle. We notice that:

- Since [tex]\theta[/tex] and v constantly change, T must change as well

- At [tex]\theta=0^{\circ}[/tex] (equilibrium position), [tex]cos \theta=1[/tex] (maximum value), and also the speed v is maximum, so the tension has the maximum value at the equilibrium position

- For [tex]\theta[/tex] increasing, the [tex]cos \theta[/tex] decreases and the speed v decreases as well, so the tension T decreases: this means that the value of the tension will be minimum in the extreme positions.

So the correct answer is D. None of the above

Answer:

see attachment

Explanation:

The electric potential varies inversely with the distance. So, the third charge should be placed at a distance of 3 cm from the origin on the x-axis.

The work done on an electric charge to shift it from infinity to a point is known as electric potential at that point. And its expression is,

[tex]V = \dfrac{kq}{r}[/tex]

here, k is the coulomb's constant.

Given data:

The magnitude of two point charges are, [tex]+2.0 \;\rm \mu C[/tex]  and  [tex]-6.0 \;\rm \mu C[/tex].

The location of each charge on the x-axis is -1.0 cm and +2.0 cm.

Let the third charge ( [tex]+3.0 \;\rm \mu C[/tex] ) be placed at a distance of x. Then the electric potential at origin is,

[tex]V = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}[/tex]

Since, potential at origin is zero (V = 0). Then,

[tex]0 = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}\\\\\dfrac{k \times 6.0}{0.02} = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times 3.0}{x}\\\\\dfrac{6.0}{0.02} = \dfrac{2.0}{0.01} +\dfrac{3.0}{x}\\\\x = 0.03 \;\rm m =3 \;\rm cm[/tex]

Thus, we can conclude that the third charge should be placed at a distance of 3 cm from the origin on the x-axis.

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Answer:

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

[tex]s=\frac{1}{2}at^2[/tex]

where

s is the vertical distance covered

a is the acceleration due to gravity

t is the time

On this celestial body, the equation is

[tex]s=10.04 t^2[/tex]

this means that

[tex]\frac{1}{2}g = 10.04[/tex]

so the acceleration of gravity on the body is

[tex]g=2\cdot 10.04 = 20.08 m/s^2[/tex]

The velocity of an object in free fall starting from rest is given by

[tex]v=gt[/tex]

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

is

[tex]t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s[/tex]

It takes approximately [tex]\( 1.424 \)[/tex] seconds for the rock to reach a velocity of 28.6 m/s on this celestial body.

We need to use the given equation and the relationship between position, velocity, and acceleration.

The equation for the position [tex]\( s \)[/tex] as a function of time [tex]\( t \)[/tex] is given by:

[tex]\[s = 10.04t^2\][/tex]

Step 1: Find the Acceleration

This equation is similar to the general form of the kinematic equation for free fall under constant acceleration:

[tex]\[s = \frac{1}{2} a t^2\][/tex]

Comparing the two equations:

[tex]\[10.04t^2 = \frac{1}{2} a t^2\][/tex]

We can solve for the acceleration [tex]\( a \)[/tex]:

[tex]\[10.04 = \frac{1}{2} a\][/tex]

[tex]\[a = 2 \times 10.04 = 20.08 \, \text{m/s}^2\][/tex]

Step 2: Use the Acceleration to Find the Time

The velocity [tex]\( v \)[/tex] of an object in free fall under constant acceleration is given by:

[tex]\[v = at\][/tex]

We need to find the time [tex]\( t \)[/tex] when the velocity [tex]\( v \)[/tex] is 28.6 m/s:

[tex]\[28.6 = 20.08 t\][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[t = \frac{28.6}{20.08}\][/tex]

[tex]\[t \approx 1.424 \, \text{seconds}\][/tex]

Answer:

Work done, W = 5941 joules

Explanation:

It is given that,

Force exerted on the block, [tex]F=(200i-149j)\ N[/tex]

Displacement, [tex]x=(23i-9j)\ m[/tex]

Let W is the work done by the force do on the block during the displacement. Its formula is given by :

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

[tex]W=(200i-149j){\cdot} (23i-9j)[/tex]                    

Since, i.i = j.j = k.k = 1

[tex]W=4600+1341[/tex]

W = 5941 joules

So, the work done by the force do on the block during the displacement is 5941 joules. Hence, this is the required solution.

Work is the energy transferred to an object by the application of force along a displacement. The work done by the force on the block during the displacement is 5941 J.

Work is the energy transferred to an object by the application of force along a displacement.

Given Here,

Displacement d = (23 m) i - (9 m) j

Force exerted on the block  F = (200 N) i - (149 N) j  

Work formula,

W = F.d

W =  (200 N) i - (149 N) j  . (23 m) i - (9 m) j

Since i.i = j.j = k.k = 1

Hence,

W = 4600 + 1341

W = 5941 J

Hence we can conclude that the work done by the force on the block during the displacement is 5941 J.

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Answer:

(a) 2.512 m

(b) 144 m

(c) 904.32 m

Explanation:

radius, r = 4.8 m

(a) for 30 degree

As we know that in 360 degree it rotates a complete round that means circumference.

In 360 degree, it rotates = 2 x π x r

in 30 degree, it rotates = 2 x π x r x 30 / 360

                                      = 2 x 3.14 x 4.8 x 30 / 360

                                      = 2.512 m

(b) for 30 rad

As we know that in one complete rotation, it rotates by 2π radian.

so,

for 2π radian it rotates = 2 x π x r

for 30 radian, it rotates = 2 x π x r x 30 / 2 π = 144 m

(c) For 30 rev

In one complete revolution, it travels = 2 x π x r

in 30 rev, it travels = 2 x π x r x 30 = 2 x 3.14 x 4.8 x 30 = 904.32 m

The speed of the wave is 309 m/s.

The equation given is y = (0.0120 m)sin[(927 rad/s)t - (3.00 rad/m)x], where y represents the displacement of a string particle and x represents the position of the particle on the string. The wave is traveling in the +x direction. To find the speed v of the wave, we need to determine the wave velocity. The wave velocity is given by the formula v = ω/k, where ω is the angular frequency and k is the wave number.

In the equation y = (0.0120 m)sin[(927 rad/s)t - (3.00 rad/m)x], the angular frequency is 927 rad/s and the wave number is 3.00 rad/m. Therefore, the wave velocity is v = 927 rad/s / 3.00 rad/m = 309 m/s.

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, [tex]q = 5.7\ \mu C=5.7\times 10^{-6}\ C[/tex]

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

[tex]E=\dfrac{q}{A\epsilon_o}[/tex]

[tex]A=\dfrac{q}{E\epsilon_o}[/tex]

[tex]A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}[/tex]

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

(a) 0.30 m/s

The total momentum of the rifle+bullet system before the shot is zero:

[tex]p_i = 0[/tex]

The total momentum of the system after the shot is the sum of the momenta of the rifle and of the bullet:

[tex]p_f = m_r v_r + m_b v_b[/tex]

where we have

[tex]m_r = \frac{W}{g}=\frac{35 N}{9.8 m/s^2}=3.57 kg[/tex] is the mass of the rifle

[tex]v_r[/tex] is the final velocity of the rifle

[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet

[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet

Since the total momentum must be conserved, we have

[tex]p_i = p_f[/tex]

So

[tex]m_r v_r + m_b v_b=0[/tex]

and so we can find the recoil velocity of the rifle:

[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{3.57 kg}=-0.30 m/s[/tex]

And the negative sign means it travels in the opposite direction to the bullet: so, the recoil speed is 0.30 m/s.

(b) 0.016 m/s

The mass of the man is equal to its weight divided by the acceleration of gravity:

[tex]m=\frac{W}{g}=\frac{650 N}{9.8 m/s^2}=66.3 kg[/tex]

This time, we have to consider the system (man+rifle) - bullet. Again, the total momentum of the system before the shot is zero:

[tex]p_i = 0[/tex]

while the total momentum after the shot is

[tex]p_f = m_r v_r + m_b v_b[/tex]

where this time we have

[tex]m_r = 66.3 kg+3.57 kg=69.9 kg[/tex] is the mass of the rifle+person

[tex]v_r[/tex] is the final velocity of the man+rifle

[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet

[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet

Since the total momentum must be conserved, we have

[tex]m_r v_r + m_b v_b=0[/tex]

and so we can find the recoil velocity of the man+rifle:

[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{66.9 kg}=-0.016 m/s[/tex]

So the recoil speed is 0.016 m/s.

The recoil speed of the rifle is 0.0031 m/s when held loosely away from the shoulder. When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg, resulting in a recoil speed of 0 m/s.

To calculate the recoil speed of the rifle in m/s, we use the principle of conservation of momentum. The momentum of the rifle before firing is equal to the momentum of the bullet after firing. The momentum of an object is calculated by multiplying its mass by its velocity. Given that the mass of the bullet is 4.5 g (0.0045 kg) and the velocity is 240 m/s, we can find the momentum of the bullet. Then, using the principle of conservation of momentum, we can calculate the recoil speed of the rifle.

(a) The momentum of the bullet is calculated as:

Momentum = mass x velocity = 0.0045 kg x 240 m/s = 0.108 kg·m/s

Since the momentum of the bullet before firing is equal to the momentum of the rifle after firing, we can write:

0.108 kg·m/s = mass of the rifle x recoil speed of the rifle

Rearranging the equation, we can solve for the recoil speed of the rifle:

Recoil speed of the rifle = 0.108 kg·m/s ÷ mass of the rifle = 0.108 kg·m/s ÷ 35 N = 0.0030857 m/s

(b) When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg. To find the recoil speed of the man and rifle together, we can again use the principle of conservation of momentum. The initial momentum of the rifle-man system is zero, as they are at rest. Therefore, the final momentum of the system after firing must also be zero. We can write:

0 = (mass of the rifle + mass of the man) x recoil speed of the system

Rearranging the equation, we can solve for the recoil speed of the system:

Recoil speed of the system = 0 ÷ (mass of the rifle + mass of the man) = 0 ÷ (28 kg + 650 N ÷ 9.8 m/s²) = 0 m/s

Answer:

[tex]8.0\cdot 10^5 N/C[/tex]

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

[tex]E=k\frac{q}{(R+r)^2}[/tex]

where

[tex]k=8.99\cdot 10^9 N m^2C^{-2}[/tex] is the Coulomb's constant

[tex]q=2.0 \mu C=2.0 \cdot 10^{-6}C[/tex] is the charge on the sphere

[tex]R=10 cm = 0.10 m[/tex] is the radius of the sphere

[tex]r=5.0 cm = 0.05 m[/tex] is the distance from the surface of the sphere

Substituting, we find

[tex]E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C[/tex]

The magnitude of the electric field due to this sphere at the given distance is 8 x 10⁵ N/C.

The given parameters;

The magnitude of the electric field due to this sphere at the given distance is calculated using Coulomb's law;

[tex]E = \frac{kQ}{R^2}[/tex]

where;

R = (0.1 + 0.05) m = 0.15 m

[tex]E = \frac{8.99\times 10^{9} \times 2\times 10^{-6}}{(0.15)^2} \\\\ E= 8 \times 10^{5} \ N/C[/tex]

Thus, the magnitude of the electric field due to this sphere at the given distance is 8 x 10⁵ N/C.

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Answer:

The kinetic energy is 135183.99 J

Explanation:

Given that,

Mass = 1158 kg

Velocity = 55 km/h = 15.28 m/s

We need to calculate the kinetic energy

The kinetic energy is equal to the half of the product of the mass and square of velocity.

Using formula of kinetic energy

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

[tex]K.E=\dfrac{1}{2}\times1158\times15.28^2[/tex]

[tex]K.E=135183.99\ J[/tex]

Hence, The kinetic energy is 135183.99 J

Answer:

The centripetal acceleration is 30 m/s²

Explanation:

Given that,

Radius = 0.30 m

Acceleration = 2.0 rad/s²

Time = 5.0

We need to calculate the centripetal acceleration

Using formula of angular velocity

[tex]\omega=a\times t[/tex]

Where,

a = acceleration

t = time

Put the value into the formula

[tex]\omega=2.0\times5.0[/tex]

[tex]\omega=10\ rad/s[/tex]

Using formula of centripetal acceleration

[tex]a_{c}=r\omega^2[/tex]

[tex]a_{c}=0.30\times10^2[/tex]

[tex]a_{c}=30\ m/s^2[/tex]

Hence, The centripetal acceleration is 30 m/s²

The centripetal acceleration of a point on the outer edge of the tire after 5.0 seconds is [tex]\( 30 \, \text{m/s}^2 \)[/tex].

The centripetal acceleration of a point on the outer edge of the tire after 5.0 seconds is given by the formula [tex]\( a_c = \omega^2 r \)[/tex], where [tex]\( \omega \)[/tex] is the angular velocity and [tex]\( r \)[/tex] is the radius of the tire.

First, we need to find the angular velocity at 5.0 seconds. Since the tire starts from rest and accelerates at a constant rate.

We can use the formula [tex]\( \omega = \omega_0 + \alpha t \)[/tex], where [tex]\( \omega_0 \)[/tex] is the initial angular velocity (0 rad/s in this case, since the tire starts from rest), [tex]\( \alpha \)[/tex] is the angular acceleration ([tex]2.0 rad/s^2[/tex]), and [tex]\( t \)[/tex] is the time (5.0 s).

Plugging in the values, we get:

[tex]\( \omega = 0 + (2.0 \, \text{rad/s}^2)(5.0 \, \text{s}) = 10.0 \, \text{rad/s} \)[/tex]

Now that we have the angular velocity, we can calculate the centripetal acceleration:

[tex]\( a_c = \omega^2 r = (10.0 \, \text{rad/s})^2(0.30 \, \text{m}) \)[/tex]

[tex]\( a_c = 100 \, \text{s}^{-2} \times 0.30 \, \text{m} \)[/tex]

[tex]\( a_c = 30 \, \text{m/s}^2 \)[/tex]

Answer: [tex]4.487(10)^{11}m[/tex]

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  

This law states a relation between the orbital period [tex]T[/tex] of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size [tex]a[/tex] of its orbit:

[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)  

Where:

[tex]T=3.87Earth-years=122044320s[/tex] is the period of the orbit of the exoplanet (considering [tex]1Earth-year=365days[/tex])

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]  

[tex]M=3.59(10)^{30}kg[/tex] is the mass of the star

[tex]a[/tex] is orbital radius of the orbit the exoplanet describes around its star.

Now, if we want to find the radius, we have to rewrite (1) as:

[tex]a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}[/tex] (2)  

[tex]a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}[/tex] (3)  

Finally:

[tex]a=4.487(10)^{11}m[/tex] This is the radius of the exoplanet's orbit

Given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.

Given the data in the question;

To determine the radius of the exoplanet's orbit, we use the equation from Kepler's Third Law:

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

Where, T is the period of the orbit of the exoplanet, G is the Gravitational Constant, M is the mass of the star and r is orbital radius.

We make "r", the subject of the formula

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

We substitute our given values into the equation

[tex]r = \sqrt[3]{\frac{(122044320s)^2*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} } \\\\r = \sqrt[3]{\frac{(1.48948*10^{16}s^2)*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} }\\\\r = \sqrt[3]{\frac{3.5689*10^{36}m^3}{4*\pi ^2} }\\\\r = \sqrt[3]{9.04*10^{34}m^3}\\\\r = 4.488*10^{11}m[/tex]

Therefore, given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.

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(a) 26.1 Nm

The magnitude of the torque exerted by a force acting perpendicularly to a surface is given by:

[tex]\tau = Fr[/tex]

where

F is the magnitude of the force

r is the distance from the pivot

In this situation,

F = 47.5 N is the force applied

[tex]r=\frac{1.10 m}{2}=0.55 m[/tex] is the distance from the hinges (the force is applied at the center of the door)

So, the magnitude of the torque is

[tex]\tau = (47.5 N)(0.55 m)=26.1 Nm[/tex]

(b) 52.3 Nm

In this case, the force is applied at the edge of the door farthest from the hinges. This means that the distance from the hinges is

r = 1.10 m

So, the magnitude of the torque is

[tex]\tau =(47.5 N)(1.10 m)=52.3 Nm[/tex]

The magnitude of torque applied about the axis of the door when the force is applied at the center is 26.125 N.m

The magnitude of the torque applied at the edge farthest from the center  is 52.25 N.m.

The given parameters;

The torque applied by the janitor is the product of applied force and perpendicular distance.

τ = F.r

The torque applied about the axis of the door when the force is applied at the center.

[tex]\tau = F \times \frac{r}{2} \\\\\tau = 47.5 \times \frac{1.1}{2} \\\\\tau = 26.125 \ N.m[/tex]

The magnitude of the torque applied at the edge farthest from the center ;

[tex]\tau = F\times r\\\\\tau = 47.5 \times 1.1\\\\\tau = 52.25 \ N.m[/tex]

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Answer:

T = 2010 N

Explanation:

m = mass of the uniform beam = 150 kg

Force of gravity acting on the beam at its center is given as

W = mg

W = 150 x 9.8

W = 1470 N

T = Tension force in the wire

θ = angle made by the wire with the horizontal =  47° deg

L = length of the beam

From the figure,

AC = L

BC = L/2

From the figure, using equilibrium of torque about point C

T (AC) Sin47 = W (BC)

T L Sin47 = W (L/2)

T Sin47 = W/2

T Sin47 = 1470

T = 2010 N

The tension in the cable supporting a 150 kg beam can be calculated by equating the weight of the beam to the vertical component of the tension in the cable. Solve 150 kg * 9.8 m/s² = T * sin(47°) for the tension T to find the force exerted by the cable.

The question concerns the calculation of the tension in the cable supporting a uniform beam. We begin by recognizing that this is a static situation with a beam of mass 150 kg in equilibrium. Thus, the total of the forces in the vertical direction must equal zero.

The forces acting on the beam are its weight (downwards) and the vertical component of the tension in the cable (upwards). The weight of the beam can be calculated by multiplying its mass by gravity g (approximately 9.81 m/s²), resulting in a downward force of 1471.5 N. The vertical component of the tension can be calculated using the sine function: T_vertical = T * sin(θ), where T is the total tension in the cable.

By setting the weight equal to the vertical component of the tension, we can solve for T: 150 kg * 9.8 m/s² = T * sin(47°). Solve this equation for the tension T to find the magnitude of the force exerted by the cable. This tension, along with its vertical and horizontal components, help maintain the beam in its horizontal position.

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Answer:

option (d)

Explanation:

E1 = 1.5 eV = 1.5 x 1.6 x 10^-19 J, E2 = 5 eV = 5 x 1.6 x 10^-19 J, c = 3 x 10^8 m/s, h = 6.62 x 10^-34 Js

Wavelength associated with 1.5 eV is λ1.

E1 = h c / λ1

λ1 = h c / E1

λ1 = (6.62 x 10^-34 x 3 x 10^8) / (1.5 x 1.6 x 10^-19)

λ1 = 8.275 x 10^-7 m = 827 nm

Wavelength associated with 5 eV is λ2.

E2 = h c / λ2

λ2 = h c / E2

λ2 = (6.62 x 10^-34 x 3 x 10^8) / (5 x 1.6 x 10^-19)

λ2 = 2.4825 x 10^-7 m = 248 nm

So, the longest wavelength is 827 nm

Answer:

+7.0 m/s

Explanation:

Let's take rightward as positive direction.

So in this problem we have:

a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)

t = 4 s time interval

v = -3.0 m/s is the final velocity (negative because it is leftward)

We can use the following equation:

v = u + at

Where u is the initial velocity

We want to find u, so if we rearrange the equation we find:

[tex]u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s[/tex]

and the positive sign means the initial direction was rightward.

Answer: [tex]1.893(10)^{27}kg [/tex]

Explanation:

This problem can be solved by the Third Kepler’s Law of Planetary motion, which states:

“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law stablishes a relation between the orbital period [tex]T[/tex] of a body (moon, planet, satellite) orbiting a greater body in space with the size [tex]a[/tex] of its orbit.

This Law is originally expressed as follows:

[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex]    (1)

Where;

[tex]T=7.16days=618624s[/tex]  is the period of the orbit Ganymede describes around Jupiter

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

[tex]M[/tex] is the mass of Jupiter  (the value we need to find)

[tex]a=10.7(10)^{8}m[/tex]  is the semimajor axis of the orbit Ganymede describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find [tex]M[/tex], we have to express equation (1) as written below and substitute all the values:

[tex]M=\frac{4\pi^{2}}{GT^{2}}a^{3}[/tex]    (2)

[tex]M=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(618624s)^{2}}(10.7(10)^{8}m)^{3}[/tex]    (3)

Finally:

[tex]M=1.8934(10)^{27}kg[/tex]   This is the mass of Jupiter

We have that for the Question  it can be said that  the magnitude of the force exerted by the horizontal rope on her arms and the ratio of the Force to the weight is

From the question we are told

A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. F-rope W =

Generally the equation for the force applied  is mathematically given as

[tex]F=\frac{( mv^2)}{R}\\\\Therefore\\\\F=\frac{( mv^2)}{R}\\\\F=\frac{( (64)(4.0)^2)}{0.890}\\\\[/tex]

F=1150.561N

b)

Generally the equation for the Weight  is mathematically given as

W=mg

Therefore

W=64*9.81

W=627.84N

Therefore

The Force to weight ratio is

[tex]F/W=1150.561N/627.84N[/tex]

F/W=1.8325

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The force exerted by the rope on the skater's arms as she moves in a circular path is 1.167 kN. This force is about 1.860 times her weight, which is 627.2 N.

The skater is experiencing centripetal force exerted by the rope, which causes her to move in a circular path. The magnitude F of this force can be calculated using the formula F = mv²/r, where m is the skater's mass (64.0 kg), v is her velocity (4.04 m/s), and r is the radius of her circular path (0.890 m).

By substituting the given numbers into this formula, we get: F = (64.0 kg)(4.04 m/s)² / 0.890 m = 1166.67 N. In kilonewtons, this force is 1.167 kN.

To compare this force with her weight, we can calculate the weight (W) using the formula W = mg, where g is the acceleration due to gravity (around 9.8 m/s²). Substituting the given mass into this formula gives us: W = (64.0 kg)(9.8 m/s²) = 627.2 N.

Comparing these two forces shows that the force exerted by the rope on her arms is about 1.860 times her weight.

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Answer:

The answer would be "His moment of inertia is 4 times the girl's"

The statement true about the boy's moment of inertia with respect to the axis of rotation  is "His moment of inertia is 4 times the girl's".

The moment of inertia is the amount of rotation obtained by an object when it is in state of motion or rest.

A boy and a girl are riding on a merry-go-round that is turning. The boy is twice as far as the girl from the merry-go-round's center. The boy and girl are of equal mass.

Moment of inertia is given by I = mr²where m is the mass and r is the distance from the axis of rotation.

For girl, I = mr²

For boy with twice the distance from axis

I = m(2r)²I = 4mr²

On comparison, we have The boy's moment of inertia is 4 times less than the girl's.

Thus, the statement true about the boy's moment of inertia with respect to the axis of rotation  is "His moment of inertia is 4 times the girl's".

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Answer:

Speed of car, v₁ = 55 m/s

Explanation:

It is given that,

Mass of Semi, m₁ = 9565 kg

Initial velocity of semi, u₁ = 55 m/s

Mass of car, m₂ = 992 kg

Initial velocity of car, u₂ = 0 (at rest)

Since, the collision between two objects is elastic and all the momentum is transferred to the car i.e final speed of semi, v₂ = 0

Let the speed of the car is v₁. Using conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]9565\ kg\times 55\ m/s+992\ kg\times 0=9565\ kg\times v_1+0[/tex]

v₁ = 55 m/s

Hence, the car move forward with a speed of 55 m/s.

Explanation:

You need to find the least common multiple (LCM) of 30 and 50.  First, write the prime factorization of each:

30 = 2×3×5

50 = 2×5²

The LCM must contain all the factors of both, so:

LCM = 2×3×5²

LCM = 150

It will take 150 minutes (or 2 hours and 30 minutes) before two trains depart at the same time again.

The next time that two trains, one on each line, will depart from Center Station at the same time is at 10:30 A.M. This is calculated by finding the least common multiple of the two train schedules.

This question requires finding the least common multiple (LCM) of the two train schedules, which represents the time duration when both trains will depart again at the same time. The LCM of 30 and 50 is 150 minutes. This means, from 8:00 A.M., it will be the next 150 minutes or 2 hours and 30 minutes when both trains will depart from Center Station at the same time. Therefore, the answer is 10:30 A.M.

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Answer:

[tex]9.96\cdot 10^{-10}J[/tex]

Explanation:

The capacitance of the parallel-plate capacitor is given by

[tex]C=\epsilon_0 k \frac{A}{d}[/tex]

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

[tex]A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2[/tex] is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

[tex]C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F[/tex]

Now we can calculate the energy of the capacitor, given by:

[tex]U=\frac{1}{2}CV^2[/tex]

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

[tex]U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J[/tex]

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

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

[tex]\tau=82.0\times0.150[/tex]

[tex]\tau=12.3\ N-m[/tex]

Now, The angular acceleration

[tex]\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}[/tex]

[tex]\dfrac{d\omega}{dt}=20.73\ rad/s^2[/tex]

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

[tex]\tau=I\times\dfrac{d\omega}{dt}[/tex]

[tex]I=\dfrac{I}{\dfrac{d\omega}{dt}}[/tex]

[tex]I=\dfrac{12.3}{20.73}[/tex]

[tex]I= 0.593\ kg-m^2[/tex]

Hence, The moment of inertia of the wheel is 0.593 kg-m².

Answer:

0.593 kg-m²

Explanation:

edg.

Answer:

Induced emf, [tex]\epsilon=2.2\ mV[/tex]

Explanation:

It is given that,

Rate of change of magnetic field, [tex]\dfrac{dB}{dt}=0.23\ T/s[/tex]

A UHF antenna is oriented at an angle of 47° to a magnetic field, θ = 47°

Diameter of the antenna, d = 13.4 cm

Radius, r = 6.7 m = 0.067 m

We need to find the induced emf of the antenna. It is given by :

[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]

Where

[tex]\phi[/tex] = magnetic flux, [tex]\phi=BA\ cos\theta[/tex]

So, [tex]\epsilon=\dfrac{d(BA\ cos\theta)}{dt}[/tex]

B = magnetic field

[tex]\epsilon=A\dfrac{d(B)}{dt}\ cos\theta[/tex]

[tex]\epsilon=\pi r^2\times \dfrac{dB}{dt}\times cos(47)[/tex]

[tex]\epsilon=\pi (0.067\ m)^2\times 0.23\ T-s\times cos(47)[/tex]

[tex]\epsilon=0.0022\ V[/tex]

[tex]\epsilon=2.2\ mV[/tex]

So, the induced emf of the antenna is 2.2 mV. Hence, this is the required solution.

Answer:

29.0 g

Explanation:

The mass of the piece of gold is given by:

m = dV

where

m is the mass

d is the density

V is the volume of the piece of gold

The density of gold is

d = 19.3 g/cm^3

while the volume of the sample is equal to the volume of displaced water, so

V = 64.5 mL - 63.0 mL = 1.5 mL

And since

1 mL = 1 cm^3

the volume is

V = 1.5 cm^3

So the mass of the piece of gold is:

m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g

Answer:

9.47 rad/s^2

Explanation:

Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,

s = 35.4 m

let a be the linear acceleration.

Use III equation of motion.

v^2 = u^2 + 2 a s

7.1 x 7.1 = 0 + 2 x a x 35.4

a = 0.71 m/s^2

Now the relation between linear acceleration and angular acceleration is

a = r x α

where,  α is angular acceleration

α = 0.71 / 0.075 = 9.47 rad/s^2

The electrostatic potential energy if all of the charges are negative is

U = -1.026 x [tex]10^{-5}[/tex] J.

We have,

The electrostatic potential energy U of a system of point charges can be calculated using the formula:

U = 1/4π∈ [tex]\sum_{i = 1}^n \sum_{j > i}^n q_iq_j/r_{ij}[/tex]

Now,

n is the number of charges.

q are the magnitudes of charges.

r is the distance between the charges.

∈ is the vacuum permittivity

∈ = 8.85 x [tex]10^{-12}[/tex] C² / N - m²

Now,

Given that all charges are negative.

[tex]q_i[/tex] = - 3.6 mu

And they are at the corners of a square with a side of 4m, the distances between charges are all diagonals of the square is [tex]r_{ij} = 4\sqrt{2}m[/tex]

Substituting the values.

U = 1/4π∈ [tex]\sum_{i = 1}^n \sum_{j > i}^n q_iq_j/r_{ij}[/tex]

U = -1.026 x [tex]10^{-5}[/tex] J

Thus,

The electrostatic potential energy if all of the charges are negative is

U = -1.026 x [tex]10^{-5}[/tex] J.

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