Two objects are moving in the x, y plane as shown. If a net torque of 44 N∙m acts on them for 5.0 seconds, what is the change in their angular momentum?

Answer:

0.344 m/s^2

Explanation:

The first step is to calculate V

V= distance /time

= 2/1.45

= 1.379 m/s

Therefore the acceleration of the block can be calculated as follows

a= V - U/2d

= 1.379 - 0/2×2

= 1.379/4

= 0.344 m/s^2

Answer:

Rₓ = 1562.5 Ω

Explanation:

The formula for the wheat stone bridge in balanced condition is given as follows:

R₁/R₂ = R₃/Rₓ

where,

Rₓ = Unknown Resistance = ?

R₃ = 2500 Ω

R₂/R₁ = 0.625

R₁/R₂ = 1/0.625 = 1.6

Therefore,

1.6 = 2500 Ω/Rₓ

Rₓ = 2500 Ω/1.6

Rₓ = 1562.5 Ω

Answer:

Vf = Vi + at

t = Vf - Vi = 0.0m/s - 1.75 m/s

    --------- = ------------------------- = 8.8s

         a              -0.20 m/s(squared)

Explanation:

This is what I got I am pretty sure it's correct but correct me if I'm wrong : )

Answer:

a and c

Explanation:

Answer:

Energy lost = 55.28Joules

Explanation:

Using the law of momentum:

m1u1+m2u2 = m1v1+m2v2

m1 and m2 are the masses of the objects

u1 and u2 are the initial velocities.

v1 and v2 are final velocities

Given

m1 = 5kg

m2 = 4kg

u1 = 12m/s

u2 = 2m/s

v1 = 10m/s

v2 = ?

Substitute

5(12)+4(2)=4(10)+5v2

60+8 = 40+5v2

68-40 = 5v2

28 = 5v2

v2 = 28/5

v2 = 5.6m/sec

Hence the speed of the 5kg block is 5.6m/s

Energy lost = Kinetic energy after collision - kinetic energy before collision

KE after collision = 1/2m1v1²+1/2m2v2²

KE after collision = 1/2(5)10²+1/2(4)(5.6)²

= 250+62.72

= 312.72Joules

KE before collision = 1/2m1u1²+1/2m2u2²

KE before collision = 1/2(5)12²+1/2(4)(2)²

= 360+8

= 368Joules

Energy lost = 368-312.72

Energy lost = 55.28Joules

Answer:

1. force applied southward = -4 j

force applied eastward = 5 i

total force applied = 5i - 4j

magnitude of total force applied = √(5)²+(-4)²

magnitude of total force applied = √25 + 16 = √41

magnitude of total force applied = 6.4N

But the beach ball also weighs 4 N,

which means that a force of 4N is required to overcome the inertia of the ball

Net force applied on the ball = total force applied - force applied by inertia

Net force applied on the ball = 6.4 - 4

Net force applied on the ball = 2.4 N

Mass of the ball:

Mass of the ball = weight of the ball / gravitational constant

Mass of the ball = 4 / 9.8 = 0.4 kg

Acceleration of Ball:

from newton's second law of motion:

F = ma

replacing the variables

2.4 = 0.4 * a          (where a is the acceleration of the ball)

a = 2.4/0.4

a = 6 m/s²

2. Mass of astronaut = 70 kg

Weight of Earth:

Weight = Mass * acceleration due to gravity

Weight = 70 * 9.8

Weight = 686 N

Weight on Moon:

Weight = Mass * acceleration due to gravity

Weight = 70 * 1.6                             (we are given that g = 1.6 on moon)

Weight = 112 N

Weight on Venus:

Weight = Mass * acceleration due to gravity

Weight = 70 * 18.7                         (we are given that g = 18.7 on Venus)

Weight = 1309 N

Answer:

Simple enough that it seems like a trick question: gravity.

Answer:

The tension in the rope is 588 N

Explanation:

I have attached a free body diagram of the problem.

First we calculate the force applied downwards by the person's weight:

[tex]F_{weight} =m*a = 80 kg * 9.8 \frac{m}{s^{2}} = 784 N[/tex]

If the scale reads a 25% of the person' weight:

[tex]F_{Scale} = F_{weight}*0.25=784N*0.25=196N[/tex]

which is the same value (but opposite) to the effective weight of the persons ([tex]F_{Person}=-F_{Scale}[/tex])

The other 75% of the total weight is the force of the rope pulling the person upwards.

[tex]F_{Pull}= F_{weight}*0.75=784N*0.75=588N[/tex]

To verify, we can use 1st Newton's law

∑F = 0

[tex]F_{Rope}-F_{Pull}+F_{Person}-F_{Scale}=0 \\588N-588N+196N-196N=0[/tex]

Answer:

μ = 0.2041

Explanation:

Given

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

Required

Determine the coefficient of friction (μ)

The acceleration is negative because it is decreasing.

Solving further, we have that:

[tex]F_f =[/tex] μmg

Where

μ = Coefficient of kinetic friction

[tex]F_f = ma[/tex]

So, we have:

ma = μmg

Divide both sides by m

a = μg

Substitute values for a and g

2 = μ * 9.8

Solve for μ

μ = 2/9.8

μ = 0.2041

Answer:

Net force = 419.5N

Explanation:

Given the following data;

Mass = 50kg

Radius = 15m

Time = 2.1 secs

Turns = 1/4 = 0.25

In order to find the net force, we would first of all solve for the speed and acceleration of the halfback.

To find speed;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the equation;

[tex]Speed = \frac{distance}{time}[/tex]

But the distance traveled is given by the circumference of a circle = [tex] 2\pi r[/tex]

Since he covered 1/4th of a turn;

[tex] Distance = 0.25 * 2 \pi *r[/tex]

Substituting into the equation;

[tex] Speed, v = \frac {(0.25*2*3.142 * 15)}{2.1}[/tex]

[tex] Speed, v = \frac {23.565}{2.1}[/tex]

Speed, v = 11.22m/s

To find acceleration;

[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]

Where, v = 11.22m/s and r = 15m

Substituting into the equation, we have;

[tex] Acceleration, a = \frac {11.22^{2}}{15}[/tex]

[tex] Acceleration, a = \frac {125.8884}{15}[/tex]

Acceleration, a = 8.39m/s²

To find the net force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

Substituting into the equation, we have;

[tex] F = 50 * 8.39[/tex]

F = 419.5N

Therefore, the net force acting upon the halfback is 419.5 Newton.

Answer:

t = 9.25 s

Explanation:

Given that,

Initial angular velocity, [tex]\omega_o=0[/tex] (at rest)

Final angular velocity, [tex]\omega_o=0.25\ rad/s[/tex]

Angular acceleration, [tex]\alpha =0.027\ rad/s^2[/tex]

We need to find the time it take the wheel to come up to operating speed. We know that the angular acceleration in terms of angular speed is given by :

[tex]\alpha =\dfrac{\omega_f-\omega_o}{t}\\\\t=\dfrac{\omega_f-\omega_o}{\alpha }\\\\t=\dfrac{0.25-0}{0.027}\\\\t=9.25\ s[/tex]

So, it will reach up to the operating speed in 9.25 s.

Answer:

When the ball is on the frictionless side of the track , the angular speed is 89.7 rad/s.

Explanation:

Consider the ball is a solid sphere of radius 3.8 cm and mass 0.14 kg .

Given ,  mass, m=0.14 kg

Ball is released from rest at a height of, h= 0.83 m

Solid sphere of radius, R = 3.8 cm

                                       =0.038 m

From the conservation of energy

                          ΔK  = ΔU  

                [tex]\frac{1}{2}mv^2 +\frac{1}{2} I\omega^2=mgh[/tex]

Here , [tex]I=\frac{2}{5} MR^2 , v= R \omega[/tex]

  [tex]\frac{1}{2}mv^2\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R^2} )=mgh[/tex]

  [tex]\frac{1}{2} [v^2+\frac{2}{5}v^2]= gh[/tex]

[tex]\frac{7}{10} v^2=gh[/tex]

[tex]0.7v^2=gh[/tex]

 v=[tex]\sqrt{[gh/(0.7)][/tex]

=[tex]\sqrt{ [(9.8 m/s^2)(0.83 m) / (0.7) ][/tex]

= 3.408 m/s

Hence, angular speed when it is on the frictionless side of the track,

[tex]\omega=\frac{v}{R}[/tex]

   = (3.408 m/s)/(0.038 m)

[tex]\omega[/tex]   =  89.7 rad/s

Hence , the angular speed is 89.7 rad/s

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

where

f is the force

m is the mass

From the question we have

[tex]a = \frac{12}{4} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

Larger in magnitude

Explanation:

As the vectors get closer reducing the 90 degree angle, based on the "parallelogram rule" for vector addition, the magnitude of the resultant vector gets larger than when they are forming a 90 degree angle.

Answer:

9.77 m/s

Explanation:

Since the cylinder is rolling, it will have both translational and rotational kinetic energy.

From conservation of energy, its initial mechanical energy equals its final mechanical energy.

So, K' + U' = K + U where K and U are the initial kinetic and potential energies respectively and K' and U' are the final kinetic and potential energies respectively.

So, Since the cylinder starts from rest, its initial kinetic energy K = 0, its initial potential energy = mgh where m = mass of cylinder, g = acceleration due to gravity = 9.8 m/s² and h = height of incline = 7.30 mm = 7.3 × 10⁻³ m . Its final kinetic energy K' = 1/2Iω² + 1/2mv² where I = moment of inertia of cylinder = 1/2mr² where r = radius of cylinder and v = translational velocity of cylinder. Its final potential energy U' = 0. Substituting these into the equation, we have  

K' + U' = K + U

1/2Iω² + 1/2mv² + 0 = 0 + mgh

1/2(1/2mr²)ω² + 1/2mv² = mgh

1/4mr²ω² + 1/2mv² = mgh          

1/4m(rω)² + 1/2mv² = mgh    v = rω

1/4mv² + 1/2mv² = mgh

3/4mv² = mgh

v² = 4gh/3

v = √(4gh/3)

v = √(4 × 9.8 m/s² × 7.3 × 10⁻³ m/3)

v = (√286.16/3) m/s

v = (√95.39) m/s

v = 9.77 m/s

Answer:

1.2 seconds

Explanation:

Formula for total time of flight in projectile is; T = 2u/g

Now, for the time to reach maximum height, it is gotten from Newton's first equation of motion;

v = u + gt

t is time taken to reach maximum height.

But in this case, g will be negative since motion is against gravity. Final velocity will be 0 m/s at max height.

Thus;

0 = u - gt

u = gt

t = u/g

Putting t for u/g in the equation of total time of flight, we have;

T = 2t

We are given t = 1.2

Thus, T = 2 × 1.2 = 2.4

Since total time of flight is 2.4 s and time to get to maximum height is 1.2 s, then it means time to fall from maximum height back to the students hand is; 2.4 - 1.2 = 1.2 s

Answer:

The answer is below

Explanation:

a) Using the formula:

[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=15\ rev/s,w_o=initial\ velocity=4.4\ rev/s, \\\theta=distance=60\ rev\\\\Substituting:\\\\15^2=4.4^2+2(60)\alpha\\\\2(60)\alpha=15^2-4.4^2\\\\2(60)\alpha=205.64\\\\\alpha=1.71\ rev/s^2[/tex]

b) The disk is initially at rest. Using the formula:

[tex]\theta=\omega_ot+\frac{1}{2}\alpha t^2 \\\\but\ \omega_o=0(rest), \theta=60\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\60=0+\frac{1}{2}(1.71) t^2\\\\t=8.4\ s[/tex]

c)

[tex]\omega=\omega_o+\alpha t \\\\but\ \omega_o=0(rest), \omega=4.4 \ rev/s\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\4.4=1.71t\\\\t=2.6\ s[/tex]

d)

[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=4.4\ rev/s,w_o=initial\ velocity=0\ rev/s, \\\alpha=1.71\ rev/s^2\\\\Substituting:\\\\4.4^2=0+2(1.71)\theta\\\\19.36=3.42\theta\\\\\theta=5.66\ rev[/tex]

Answer:

A force of 1.788 newtons is exerted on one side of the sheet of paper by the atmosphere.

Explanation:

From definition of pressure, we get that force exerted by the atmosphere ([tex]F[/tex]), measured in newtons, on one side of the sheet of paper is given by the expression:

[tex]F = P_{atm}\cdot w\cdot l[/tex] (Eq. 1)

Where:

[tex]P_{atm}[/tex] - Atmospheric pressure, measured in pascals.

[tex]w[/tex] - Width of the sheet of paper, measured in meters.

[tex]l[/tex] - Length of the sheet of paper, measured in meters.

If we know that [tex]P_{atm} = 101325\,Pa[/tex], [tex]w = 0.353\,m[/tex] and [tex]l = 0.05\,m[/tex], then the force exerted on one side of the sheet of paper is:

[tex]F = (101325\,Pa)\cdot (0.353\,m)\cdot (0.05\,m)[/tex]

[tex]F = 1.788\,N[/tex]

A force of 1.788 newtons is exerted on one side of the sheet of paper by the atmosphere.

Answer:

1730nm

Explanation:

Observed wavelength λ= 1.9 millimeters

Red shift of photons = 1100

Emitted wavelength = λo

Red shift = λ - λo/ λo = 1100

= [(1.9x10^-3 - λo)/ λo] = 1100

= 1.9x10^-3 - λo = 1100 λo

= 1.9x10^-3 = 1100 λo + λo

= 1.9x10^-3 = 1101 λo

We divide through by 1101 to get the value of lambda

1.9x10^-3/1101 = λo

λo = 0.000001726

λo = 1726nm

This is approximately

λo = 1730nm

Thank you!

Given :

Force on object moving with constant velocity due to magnetic field is given by :

[tex]\vec{F}=q\vec{v}\times \vec{B}\\\\\vec{F}=-5[ ( \hat{i} + 7\hat{j}) \times ( 10\hat{k})]\\\\\vec{F}=-5[ 10(-\hat{j})+70(\hat{i})]\\\\\vec{F}=50\hat{j}-350\hat{i}\ N[/tex]

Now,

[tex]F = \sqrt{50^2+350^2}\\\\F=353.55\ N[/tex]

Hence, this is the required solution.

A particle with charge q = +5.00 C that is moving with some velocity, then the magnetic force vector on the particle will be 353.55 N.

The phrase "vector" in physics and mathematics is used popularly to refer to some values that cannot be stated by a simple integer (a scalar) or to certain elements of high - dimensional space.

For things like displacements, forces, and velocity can have both a magnitude and direction, vectors were initially introduced in geometry and physics. Similar to how lengths, masses, and time are represented by real numbers, these quantities were represented by geometric vectors.

In some contexts, tuples—finite sequences of numbers with a definite length—are sometimes referred to as vectors.

From the data given in the question,

F = qv × B

F = -5 [(î +7 ĵ) ×(10 k)]

F = 50 ĵ - 350 î N

Now, calculate the force,

F = √50² + 350²

F = 353.55 N.

To know more about Vector:

https://brainly.com/question/13322477

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Explanation:

Speed - The rate at which something moves

Velocity - The speed of something in a specific direction

Velocity is kind of a specific type of speed.

Answer:

The friction force exerted by the surface is 490 newtons.

Explanation:

From Physics, we remember that static friction force ([tex]f_{s}[/tex]), measured in newtons, for a particle on a horizontal surface is represented by the following inequation:

[tex]f_{s} \leq \mu_{s}\cdot m \cdot g[/tex] (Eq. 1)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

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

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If [tex]f_{s} = \mu_{s}\cdot m \cdot g[/tex], then crate will experiment an imminent motion. The maximum static friction force is:  ([tex]\mu_{s} = 0.6[/tex], [tex]m = 100\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])

[tex]f_{s} = (0.6)\cdot (100\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]f_{s} = 588.42\,N[/tex]

From Newton's Laws we get that current force of friction as reaction to the pulling force done by the worker on the crate is:

[tex]\Sigma F = F-f = 0[/tex]

[tex]f = F[/tex] (Eq. 2)

Where:

[tex]F[/tex] - Horizontal force done by the worker, measured in newtons.

[tex]f[/tex] - Static friction force, measured in newtons.

If [tex]F = 490\,N[/tex], then the static friction force exerted by the surface is:

[tex]f = 490\,N[/tex]

Given that [tex]f < f_{s}[/tex], the crate does not change its state of motion. The friction force exerted by the surface is 490 newtons.  

Answer:

Kindly check explanation

Explanation:

Change in momentum:

m(v - u)

M = mass = 1500kg

v = final velocity = 0 m/s

u = initial velocity = 8m/s

Momentum = 1500(8 - 0)

Momentum = 1500 * 8

Momentum = 12000 kgm/s

B.) impulse that acts on the car

Impulse = Force * time

Impulse = change in momentum with time

Impulse = m(v - u)

Hence, impulse = 12000Ns

C.) Average force action on the car during collision

Impulse = Force * time

12000 = force * 0.6

12000/ 0.6 = force

Force = 20000 N

D.) Workdone by haystack in stopping the car :

Workdone = Force * Distance

Distance = speed * time

Distance = 8 * 0.6

Distance = 4.8m

Hence,

Workdone = 20,000 * 4.8 = 96000 J

Answer:

A)90°

Explanation:

ufl.edu A wire of length 27.7 cm carrying a current of 4.11 mA is to be formed into a circular coil and placed in a uniform magnetic field of magnitude 5.85 mT. The torque on the coil from the field is maximized. (a) What is the angle between and the coil's magnetic dipole moment? (b) What is the number of turns in the coil? (c) What is the magnitude of that maximum torque?

(a) What is the angle between and the coil's magnetic dipole moment.

The angle between and the coil's magnetic dipole moment is 90°, because it is perpendicular to the field since there is maximum torgue.

(b) What is the number of turns in the coil?

Answer:

kinetic

Explanation:

Answer:

kinetic energy.

Explanation:

Answer:

[tex]\alpha =240.79\ rad/s^2[/tex]

Explanation:

Given that,

Initial angular velocity, [tex]\omega_i=0[/tex]

Final angular velocity, [tex]\omega_f=16740\ rev/min = 1753\ rad/s[/tex]

We need to find the angular acceleration of the drill. It is given by the formula as follows :

[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{1753-0}{7.28}\\\\\alpha =240.79\ rad/s^2[/tex]

So, the angular acceleration of the drill is [tex]240.79\ rad/s^2[/tex].

correct answer is mass media is correct answer OK

Answer:

1.6m

Explanation:

Using the spring equation, which is as follows:

F = Kx

Where; F = force applied on the spring (N)

K = spring constant (3kg/s²)

x = extension of spring (m)

However, Force applied by object = mass × acceleration (9.8m/s²)

F = 0.49kg × 9.8m/s²

F = 4.802N

Since, F = 4.802N

F = Kx

4.802 = 3 × x

4.802 = 3x

x = 4.802/3

x = 1.6006

x = 1.6m

The spring is stretched i.e. the extension of the spring, by 1.6m