All categories

3 years ago

When grip strength increases:

Physics

1 answer:

Nitella [24]3 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. <em>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.</em>

You might be interested in

A parallel-plate capacitor is constructed of two horizontal 15.6-cm-diameter circular plates. A 1.4 g plastic bead with a charge

MrRissso [65]

Answer: The upper plate is positive charged. b) 0.53 µC.

Explanation:

As the plastic bead is suspended between the two plates, the external forces acting upon it must be equal and opposite each other.

The external forces acting upon it, are gravity (always downward) and the on due to electric Field, which must aim upwards, to keep the system in equilibrium.

As the charge on which the force must be act is a negative charge, this means that the force and the electric field must aim in opposite direction, so the electric field must point from the upper plate to the lower one.

This means that the upper plate must be positive charged.

We can find out the magnitude of the electric field, as follows:

E = mg / q = 0.0014 Kg. 9.8 m/s2 / 4.4 x 10⁻⁹ C = 3.12. 10⁶ V/m

b) In order to know the charge on the positive charged plate, we know that the capacitance , by definition, is as follows:

C = Q /V

It can be showed, that for a parallel plate capacitor, the capacitance can be expressed as a relationship of geometric parameters, as follows:

C = ε₀ A / d

But, V and E are related linearly by the expression:

V= E. d

Equating both expressions for C, we have:

Q / E.d = ε₀ A / d , so we can simplify and solve for Q, as follows:

Q = ε₀ A E = 8.85. 10⁻¹² farad/m. π (0.156)2 . ¼. 3.12. 10⁶ V/m = 0.53 µC

4 0

4 years ago

Which of the three types of equations follows the law of conservation of mass?

ArbitrLikvidat [17]

Answer:

BE MORE DESCRIPTIVE

Explanation:

8 0

3 years ago

Steam enters an adiabatic turbine at 6 MPa, 600 ℃, and 80 m/s and leaves at 50 kPa, 100 ℃, and 140 m/s. If the power output of t

lisabon 2012 [21]

Answer:

W(r,out) = 5.81 MW

$\eta$ = 86.1 %

Explanation:

we use here steam table for get value of h1, s1 etc

so use for 6MPa and 600 degree

Enthalphy of steam h1 = 3658.8 kJ/kg

Entropy of steam s 1 is = 7.1693 kJ /kg.K

and

for 50 kPa and 100 degree

Enthalphy of steam h2 = 2682.4 kJ/kg

Entropy of steam s2 is = 7.6953 kJ /kg.K

so we use here energy balance equation that is

$m\times(h1 + \frac{v1^2}{2} = m\times(h2 + \frac{v2^2}{2} + W(out)$ ..............1

put here value and we get m

m = $\frac{5\times1000}{3658.8-2682.4+\frac{80^2-140^2}{2}\times \frac{1}{1000}}$

solve it we get

m = 5.156 kg/s

so by energy balance equation

m $\psi$ 1 = m $\psi$ 2 + W(r,out)

W(r,out) = m( $\psi$ 1 - $\psi$ 2)

W(r,out) = h1 - h2 + ΔKE + ΔPE - To(s1-s2)

W(r,out) = $m[h1-h2+ \frac{v1^2-v^2}{2}- To (s1-s2)$

W(r,out) = W(a,out) - m.To.(s1-s2) ........................2

put here value

W(r,out) = 5000 - ( 5.156 × (25 + 273) ×( 7.1693 - 7.6953)

W(r,out) = 5908.19 = 5.81 MW

and

second law deficiency is

$\eta$ = $\frac{W(a,out)}{W(r,out)}$ ..............................3

put here value

$\eta$ = \frac{5}{5.81}

$\eta$ = 86.1 %

6 0

3 years ago

A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done inthe process.

dlinn [17]

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

Force= 9.80665 x 0.5kg
Force= 4.90332 Newtons

Distance changed= 5-3
distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

5 0

2 years ago

When running on its 11.4 V battery, a laptop computer uses 8.3 W. The computer can run on battery power for 8.5 h before the bat

AveGali [126]

Answer:

The laptop battery is capable of supplying 253980 J.

Explanation:

Given;

voltage of the battery, V = 11.4 V

power consumed by the laptop, P = 8.3 W

duration of the battery before depletion, t = 8.5 hours

Determine the amount of energy supplied by the laptop battery within 8.5 hours.

Energy, E = Power x time

Energy, E = 8.3 W x 8.5 h = 70.55 Wh

Energy, E in joules = 70.55 wh x 3600 s/h = 253980 J

Therefore, the amount of energy supplied by the laptop battery within 8.5 hours is 253980 J.

7 0

3 years ago