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liberstina [14]
3 years ago
10

A vertical scale on a spring balance reads from 0 to 200 \rm{N}. The scale has a length of 10.0 \rm{cm} from the 0 to 200 \rm{N}

reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.05 \rm{Hz}.
Ignoring the mass of the spring, what is the mass m of the fish?
Physics
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

m = 12.05 kg

Explanation:

Spring constant in K, N/m

K = 200/10* 100

K = 2000 N/m

Angular Frequency = sqrt (Spring constant / (Mass )

ω = 2 π f

ω =  2π* 2.05 Hz = 12.8805 rad/s

ω^2 = Spring constant / Mass

Mass= Spring constant / ω^2

ω^2 = 165.907 rad^2/s^2

m = 2000 (N/m)/165.907 (rad^2/s^2)

m = 12.05 kg

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Therese plays lacrosse, but she has injured her ankle and has to take a few weeks off. She is finding that she has a lot of trou
vivado [14]

Answer: No endorphins

Explanation: cuz' she injured her ankle

6 0
3 years ago
A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th
N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0
3 years ago
Determine the magnitude of the current flowing through a 4.7 kilo ohms resistor if the voltage across it is (a) 1mV (b) 10 V (c)
mariarad [96]

Answer:

213 nA

2.13 mA

851e^-t μA

Explanation:

We have a pretty straightforward question here.

Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as

V = IR, since we need I, we can write that

I = V/R

a) at V = 1 mV

I = (1 * 10^-3) / 4.7 * 10^3

I = 2.13 * 10^-7 A or 213 nA

b) at V = 10 V

I = 10 / 4.7 * 10^3

I = 0.00213 A or 2.13 mA

c) at V = 4e^-t

I = 4e^-t / 4.7 * 10^3

I = 0.000851e^-t A or 851e^-t μA

5 0
3 years ago
. A 1.50kg mass on a spring has a displacement as a function of time given by the equation: x(t) = (7.40cm)cos[(4.16s-1)t – 2.42
rusak2 [61]

Answer:

Solution:

we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]

Position, velocity and acceleration are all based on the equation of motion.

The equation represents the position.  The first derivative gives the velocity and the 2nd derivative gives the acceleration.

x(t)=8sint

x'(t)=8cost

x"(t)=-8sint

now at time t=2pi/3,

position, x(t)=8sin(2pi/3)=4*squart(3)cm.

velocity, x'(t)=8cos(2pi/3)==4cm/s

acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2

so at present the direction is in y-axis.

5 0
3 years ago
21. Calculate the acceleration of the bus from point D to E. Show your work.
Marat540 [252]

21) Acceleration from D to E: 1 m/s^2

22) The acceleration of the bus from D to E is 1 m/s^2

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

a=\frac{5-0}{5}=1 m/s^2

22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
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