PHYSICS COLLEGE

A vertical scale on a spring balance reads from 0 to 245 N . The scale has a length of 10.0 cm from the 0 to 245 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.55 Hz . Ignoring the mass of the spring, what is the mass m of the fish?

Answers

Answer 1

Answer:

Answer:

mass m of the fish is 7.35 kg

Explanation:

Given data

spring balance reads = 0 to 245 N

length = 10.0 cm

scale reading = 0 to 245 N

frequency f = 2.55 Hz

to find out

mass m of the fish

solution

we know the relation that is

ω = √(k/m) ......................1

here k = spring reading / length = 245 / 0.135

k = 1814.81 N/m

and

ω = 2π × f

ω = 2π × 2.5 = 15.71 rad/s

so put all value in equation 1 we get m

ω = √(k/m)

15.71 = √(1814.81/m)

so m = 7.35

mass m of the fish is 7.35 kg

Answer 2

Answer:

Answer:

9.55 kg

Explanation:

F = 245 N

Let K be te spring constant

F = K x

K = 245 / 0.1 = 2450 N/m

ω = 2 x π x f = 2 x 3.14 x 2.55 = 16.014 rad/s

$\omega =\sqrt{(K)/(m)}$

where m be the mass of fish

$16.014 =\sqrt{(2450)/(m)}$

m = 9.55 kg

Related Questions

Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).
PLEASE HELP!!! Sophia says that if we add two electrons to oxygen in order to fill its valence shell, it’s expected charge would be +2. Is she correct? If not, explain the error in her thinking.
If my weight on Earth is 140lbs, what is my mass?
A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?
Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

A car turns from a road into a parking lot and into an available parking space. The car’s initial velocity is 4 m/s [E 45° N]. The car’s velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.

Answers

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→ θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answers

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

$\omega_(f)^(2) = \omega_(0)^(2) + 2\alpha*\theta$

Where:

$\omega_(f)$ : is the final angular velocity = 0

$\omega_(0)$ : is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

$\alpha = (\tau)/(I)$

Where:

I: is the moment of inertia for the disk

τ: is the torque

The moment of inertia is:

$I = (mr^(2))/(2)$

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

$I = (105.00 kg*(0.297 m)^(2))/(2) = 4.63 kg*m^(2)$

Now, the torque is equal to:

$\tau = -F x r = -\mu*F*r$

Where:

F: is the applied force = 46.650 N

μ: is the kinetic coefficient of friction = 0.451

$\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m$

The minus sign is because the friction force is acting opposite to motion of grindstone.

Having the moment of inertia and the torque, we can find the angular acceleration:

$\alpha = (-6.25 N*m)/(4.63 kg*m^(2)) = -1.35 rad/s^(2)$

Finally, we can find the number of turns that the stone will make before coming to rest:

$0 = \omega_(0)^(2) + 2\alpha*\theta$

$\theta = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns$

I hope it helps you!

Why We can’t Cure Aging? Support your answer?

Answers

We can’t cure aging because it is inevitable. Aging is a part of life and it leads to death which is also inevitable and inescapable. We can’t stop our aging because it is natural and normal to age

A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

Answers

Answer:

93.6 N in the 41° cable
155.6 N in the 63° cable
200 N in the vertical cable

Explanation:

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

U = Tcos(41°)/cos(63°) . . . . write an expression for U

The vertical components must total 200 N, so we have ....

Tsin(41°) +Usin(63°) = 200

Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

41°: 93.6 N

63°: 155.6 N

90°: 200 N

The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.

Given the following data:

Force = 200 Newton.

Angle 1 = 41°

Angle 2 = 63°

How to calculate the tension.

First of all, we would determine the third tension force based on the vertical component as follows:

$\sum F_y = 0\n\nT_3 - F_g =0\n\nT_3 - F_g=200\;N$

Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:

For the horizontal component:

$\sum F_x = -T_1cos41+T_2cos 63=0\n\n0.7547T_1=0.4540T_2\n\nT_1=(0.4540T_2)/(0.7547)\n\nT_1 = 0.6016T_2$ ....equation 1.

For the vertical component:

$\sum F_y = T_1sin41+T_2sin 63-T_3=0\n\n\sum F_y = T_1sin41+T_2sin 63-200=0\n\n0.6561T_1+0.8910T_2 =200$ ...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

$0.6561 * (0.6016T_2)+0.8910T_2 =200\n\n0.3947T_2+0.8910T_2 =200\n\n1.2857T_2 =200\n\nT_2 = (200)/(1.2857) \n\nT_2 = 155.56\;Newton$

For the first tension:

$T_1 = 0.6061T_2\n\nT_1 = 0.6061 * 155.56\n\nT_1 = 94.29\;Newton$

Read more on tension here: brainly.com/question/4080400

. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force

Answers

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration

∑ F = m a

in this case there are two forces on the x axis

F_applied - fr = 0

since they indicate that the velocity is constant, consequently

F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it takethe ball to reach the highest point?

Answers

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

Other Questions

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?3) For the conditions, how long does it take for her to reach the opposite bank?

Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.