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2 years ago

Determine the elastic energy U stored in the compressed spring.

Physics

1 answer:

hoa [83]2 years ago

7 0

Answer:

Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.

Explanation:

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Let A be the last two digits and let B be the sum of the last three digits of your 8-digit student ID. (Example: For 20245347, A

zheka24 [161]

Answer:

The speed is 173 m/s.

Explanation:

Given that,

A = 47

B = 14

Length 1 urk = 58.0 m

An hour is divided into 125 time units named dorts.

3600 s = 125 dots

dorts = 28.8 s

Speed v= (25.0+A+B) urks/dort

We need to convert the speed into meters per second

Put the value of A and B into the speed

$v=25.0+47+14$

$v =86\ urk s/dort$

$v=86\times\dfrac{58.0}{28.8}$

$v=173.19\ m/s$

Hence, The speed is 173 m/s.

7 0

3 years ago

A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio

pickupchik [31]

Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

$m=\dfrac{W}{g}$

$m=\dfrac{110\ N}{9.8\ m/s^2}$

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

$M=d\times V$

d is the density of water, $d=1000\ kg/m^3$

V is the volume of rock, $V=0.00337\ m^3$

$M=1000\ kg/m^3\times 0.00337\ m^3$

M = 3.37 kg

The apparent weight in water, W = m - M

$W=7.85\ kg\times 9.8\ m/s^2$

W = 76.93 N

So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.

4 0

2 years ago

An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th

Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0

2 years ago

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in se

ASHA 777 [7]

Answer:

R = 9.85 ohm , r = 0.85 ohm

Explanation:

Let the two resistances by r and R.

when they are connected in series:

V = 12 V

i = 1.12 A

The equivalent resistance when they are connected in series is

Rs = r + R

So, By using Ohm's law

V = i Rs

Rs = V / i = 12 / 1.12 = 10.7 ohm

R + r = 10.7 ohm .... (1)

When they are connected in parallel:

V = 12 V

i = 9.39 A

The equivalent resistance when they are connected in parallel

$R_{p}=\frac{R+r}{rR}$

So, By using Ohm's law

V = i Rp

Rp = V / i = 12 / 9.39 = 1.28 ohm

$\frac{R+r}{rR}=1.28$ .... (2)

by substituting the value of R + r from equation (1) in equation (2), we get

r R = 8.36 ..... (3)

$R-r = \sqrt{\left ( R+r \right )^{2}-4rR}$

$R-r = \sqrt{\left ( 10.7 \right )^{2}-4\times 8.36}=9$ ..... (4)

By solvng equation (1) and (4), we get

R = 9.85 ohm , r = 0.85 ohm

8 0

3 years ago

If the mass of one of two objects is increased, the force of attraction between them will

Sonja [21]

It will increase..........................

5 0

2 years ago