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2 years ago

When grip strength increases:

Physics

1 answer:

Nitella [24]2 years ago

5 0

Answer:

e. the number of active motor units increases.

Explanation:

There is a direct relationship between the number of active motor units and the grip strength in a given scenario. For example, increase in the grip strength leads to increase in the number of active motor units. In the other-hand, the decrease in grip strength leads to the decrease in the number of active motor units.

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A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate w

grigory [225]

A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2 to the east , how far did it travel during the accelration? Give your answer in units of kilometers per hour/sec. To find the acceleration of the car we have to 
1. First determine the suitable formula for this word problem.
Which is a. A=vf-vi/t which will be
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s
Solution: 
x = v0t + ½at2
x = 16.645375 m

7 0

3 years ago

A 4.0kg block is sliding with a constant velocity of 3.0m/s on a frictionless table that is 0.5m high. If all of the block’s ene

MArishka [77]

Answer:

Velocity = 4.33[m/s]

Explanation:

The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.

$E_{M}=E_{p} + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2} \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]$

All this energy will become kinetic energy and we can find the velocity.

$37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]$

8 0

3 years ago

Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?

MArishka [77]

This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

v = 4.6 m/s
d = ¿?
t = 10 sec

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the speed is equal to 4.6 m/s, the time is equal to 10 s, which is left as follows:

$\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }$

$\bf{d=46 \ m}$

Therefore, the speed at 10 seconds is 46 meters.

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6 0

2 years ago

2. An alternating current is represented by the equation I=20sin 100mt.

Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

$I=I_o sin\omega t$

Where

I₀ is the peak value of current

$\omega$ is angular frequency

$As\ \omega=2\pi f$

So,

$f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{100\pi}{2\pi}\\\\f=50\ Hz$

We know that,

$I_{rms}=\dfrac{I_0}{\sqrt{2}}\\\\=\dfrac{20}{\sqrt{2}}\\\\I_{rms}=14.14\ A$

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.

3 0

2 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

7 0

3 years ago