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3 years ago

13

It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

e's law and is depressed 0.65 cm by its maximum load of 110 kg.

1 answer:

Molodets [167]3 years ago

7 0

<span>F = kx, where F is the force we apply, k is the spring constant, and x is the extension of the material
so the </span><span>spring's effective spring constant k is:
k = F / x
k = ( 110 kg ) ( 9.81 m/s^2) / 0.65 cm
k = 1660 N/cm is the </span>spring's effective spring constant

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Salmon often jump waterfalls to reach their

erik [133]

Answer:

6.35 m/s

Explanation:

The motion of the salmon is equivalent to that of a projectile, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with constant acceleration ( $g=-9.8 m/s^2$ , acceleration of gravity)

The horizontal velocity of the salmon is given by:

$v_x = u cos \theta$

where

u = ? is the initial speed

$\theta=32^{\circ}$ is the angle of projection

Then the horizontal distance covered by the salmon after a time t is given by

$d=v_x t =(u cos \theta) t$

Or equivalently, the time taken to cover a distance d is

$t=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$h = (u sin \theta) t + \frac{1}{2}gt^2$ (2)

where

$u sin \theta$ is the initial vertical velocity

If we substitute (1) into (2), we get:

$h = (u sin \theta) \frac{d}{cos \theta} + \frac{1}{2}g(\frac{d}{ ucos \theta})^2=d tan \theta + \frac{gd^2}{2u^2 cos^2 \theta}$

We now that in order to reach the breeding grounds, the salmon must travel a distance of

d = 2.02 m

reaching a height of

h = 0.574 m

Substituting these data into the equation and solving for u, we find the initial speed that the salmon must have:

$u =\sqrt{ \frac{gd^2}{2(h-d tan \theta) cos^2 \theta}}=\sqrt{\frac{(-9.8)(2.02)^2}{2(0.574-(2.02)(tan 32))(cos^2(32))}}=6.35 m/s$

8 0

3 years ago

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

3 years ago

A 50-ω resistor is connected to a 9.0 V battery. How much thermal energy is produced in 7.5 minutes?

Aleksandr-060686 [28]

Answer:

1.2 102j

Explanation:

because it is the most important for me

3 0

3 years ago

Use this technique to find a formula for the intensity I of a sound, in terms of the sound level β and the reference intensity I

denis23 [38]

Answer:

$I = I_o (10^{\frac{\beta}{10}})$

Explanation:

As we know that sound level is given as

$\beta = 10 Log (\frac{I}{I_o})$

so here we have

$\frac{\beta}{10} = Log(\frac{I}{I_o})$

now we can take anti log both sides

$10^{\frac{\beta}{10}} = \frac{I}{I_o}$

so we have

$I = I_o (10^{\frac{\beta}{10}})$

so above is the expression for the intensity of sound in terms of reference intensity and sound level

4 0

3 years ago

In simple harmonic motion, the speed is greatest at that point in the cycle whenA) the magnitude of the acceleration is a maximu

hram777 [196]

Answer:

C) the magnitude of the acceleration is a minimum.

Explanation:

As we know that ,the general equation of the simple harmonic motion given as

The displacement x given as

x=X sinω t

Then the velocity v will become

v= X ω cosωt

The acceleration a

a= - X ω² sinω t

The speed of the particle will be maximum when cosωt will become 1 unit.

It means that sinωt will become zero.So acceleration and displacement will be minimum.

Therefore when speed is maximum then acceleration will be minimum.

At the mean position the speed of the particle is maximum that is why kinetic energy also will be maximum and the potential energy will be minimum.

Therefore option C is correct.

4 0

3 years ago