A 54.0-kg ice skater is moving at 3.94 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 0.900 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. .931416 Correct: Your answer is correct. kN (b) Compare this force with her weight. Frope W

Answer:

74 m.

Explanation:

From the question given above, the following data were obtained:

Height (h) = 186 m

Initial velocity (u) = 12 m/s

Horizontal distance (d) =?

Next, we shall determine the time taken for the rock to get to the ground. This can be obtained as follow:

Height (h) = 186 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

H = ½gt²

186 = ½ × 9.8 × t²

186 = 4.9 × t²

Divide both side by 4.9

t² = 186/4.9

Take the square root of both side

t = √(186/4.9)

t = 6.16 s

Thus, the time taken for the rock to get to the ground is 6.16 s

Finally, we shall determine the horizontal distance travelled by the rock as follow:

Initial velocity (u) = 12 m/s

Time (t) = 6.16 s

Horizontal distance (d) =?

Horizontal distance (d) = Initial velocity (u) × Time (t)

d = u × t

d = 12 × 6.16

d = 73.92 ≈ 74 m

Therefore, the horizontal distance travelled by the rock is 74 m

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:

A OR C

Nearly A.

Explanation:

correct answer is mass media is correct answer OK

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:

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:

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:

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:

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

Explanation:dropped stone hits water:

24.4-4.9t^2=0

t=2.23

stone has fallen 6.6m:

4.9t^2 = 6.6

t = 1.16

So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water:

24+1.07v-4.9*1.07^2 = 0

v = -17.19 m/s

Both the rocks hit the water at the exact same time, and the initial velocity of Jill's rock will be -19.28 m/s.

A vector measurement of the rate and direction of motion is what is meant by the term "velocity." Simply said, velocity is the rate of movement in a single direction. Velocity may be used to gauge both the speed of a rocket launching into space and the speed of a car travelling north on a busy freeway.

According to the question, initial velocity will be 0.

By using equation of motion :

Let Jill is x and Jack is y.

s=ut+1/2at² where,

a is acceleration,

u is initial velocity and,

t is the time period.

23.4=(-1/2)(9.8)×t²

t(y)=[tex]\sqrt{(23.4)/(4.9)}[/tex]

=2.185 seconds.

Time taken by Jack's stone that it travel the first 7.2 meters.

t=[tex]\sqrt{(2*7.2)/9.8}[/tex] =1.21 seconds.

So, time taken by stone thrown by jack,

t(Jill)=t(jack)-t

=23.4=-ut(x)-1/2(gt²)(y)

-23.4=(-u×0.975)-(1/2×9.8×(0.975)²)

u= -(23.4-4.6/0.975)

u=-19.28 m/s

Hence, the initial velocity of the Jill's stone will be -19.28 m/s.

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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:

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

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:

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!

Answer:

Explanation:

angular acceleration ω = 8 rad /s²

tangential acceleration a = angular acceleration x radius of circle of rotation

For a point on the end of the rod , radius of circle = L .

tangential acceleration = 8 L .

For  a point a distance L/2 from the end of the rod, radius of circle = L /2 .

tangential acceleration = 8 L/2 = 4 L  .

It is so because angular acceleration will be same for this point , only radius changes .

required ratio = 8L / 4L

= 2.

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].

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.

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Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

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:

kinetic

Explanation:

Answer:

kinetic energy.

Explanation:

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

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:

True

Explanation:

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:

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:

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:

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 and c

Explanation: