Ask question

Ask question

All categories

3 years ago

3

The _____ tighten(s) the muscles.

2 answers:

Ierofanga [76]3 years ago

5 0

I believe it is D. The muscle spindles tighten(s) the muscles. Hope this helped.

Xelga [282]3 years ago

4 0

The answer is D: Muscle Spindles

You might be interested in

Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for l

liberstina [14]

Explanation:

Given: $\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}$

$\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}$

$\:\:\:\:\:= 6×10^{14}\:\text{Hz}$

The work function $\phi$ is then

$\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})$

$\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}$

3 0

3 years ago

Cliff divers at Acapulco jump into the sea from a cliff 37.1 m high. At the level of the sea, a rock sticks out a horizontal dis

Stella [2.4K]

Answer:

$v_x = 4.87 m/s$

Explanation:

Height of the cliff is given as

$h = 37.1 m$

now the time taken by the diver to hit the surface is given as

$h = \frac{1}{2}gt^2$

$37.1 = \frac{1}{2}(9.8)t^2$

$t = 2.75 s$

Now in the same time it has to cover a distance of 13.39 m

so the speed in horizontal direction is given as

$v_x = \frac{x}{t}$

$v_x = \frac{13.39}{2.75}$

$v_x = 4.87 m/s$

3 0

4 years ago

A 1-kilogram parcel of air is at 35°c and contains 7 grams of water vapor. What is the relative humidity?

Nitella [24]

Answer:

20%

Explanation:

Relative Humidity (%) = (water vapor content÷water vapor capacity) × 100

=(7÷35)×100

=(0.2)×100

=20%

According to the Temperature-Water Vapor Capacity Table, the water capacity at 35 °C is 35 grams.

Water Vapor Capacity: The amount of water (grams) which air can hold at a given temperature.

Water Vapor Content: The amount of water vapor actually present in the air.

8 0

3 years ago

It take my planet amount of time to orbit the sun

navik [9.2K]

It takes the Earth 365 days to orbit the sun.

4 0

4 years ago

Read 2 more answers

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the co

ladessa [460]

Answer:

$\dot W_{in} = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_{in} + \dot m \cdot c_{p}\cdot (T_{1}-T_{2}) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_{in} = \dot m \cdot c_{v}\cdot (T_{2}-T_{1})$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_{u}\cdot T$

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$

$\rho = \frac{(104\,kPa)\cdot (28.02\,\frac{kg}{kmol})}{(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (292\,K)}$

$\rho = 1.2\,\frac{kg}{m^{3}}$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,\frac{kg}{m^{3}})\cdot (0.15\,\frac{m^{3}}{s} )$

$\dot m = 0.18\,\frac{kg}{s}$

The work input is:

$\dot W_{in} = (0.18\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot K})\cdot (565\,K-292\,K)$

$\dot W_{in} = 49.386\,kW$

5 0

3 years ago