How do organisms use cellular respiration and fermentation to survive?

The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec

marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

$\epsilon = \frac{\sigma}{\sqrt{n} }$

To reduce the standard error to 0.03 s, let the additional number of trials = x

$0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials$

Therefore, the additional trials needed is 48 trials.

6 0

3 years ago

A car accelerate from a stop to 27 m/sec in 5 seconds. What was the car's acceleration?

Furkat [3]

The acceleration should 5.4 m/s^2

4 0

3 years ago

A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

alukav5142 [94]

Answer:

0.00016 kg

Explanation:

Given:

Power = P = 1.2 × 10⁹ Watts

Power = work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W

Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²

= 0.00016 kg

6 0

3 years ago

A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the

barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:

$KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i$

$k$ is Coulomb's constant.

#The electric field, $E$ at radius r is expressed as:

$E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii$

From i and ii, we have:

$KE=Eqr$

$r=(KE)/Eq$

#Substitute actual values in our equation:

$r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m$

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0

3 years ago

Read 2 more answers

A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee

SOVA2 [1]

Answer:

$\omega=0.12\frac{rad}{s}$

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

$\omega=\frac{2\pi}{T}(1)$

Here T is the period, that is, the time taken to complete onee revolution:

$T=\frac{2\pi r}{v}(2)$

Replacing (2) in (1):

$\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}$

3 0

3 years ago