Ask question

Ask question

All categories

3 years ago

9

A bug is on the rim of a 78 rev/min, 12 in. diameter record. The record moves from rest to its final angular speed in 2.67 s. Fi

nd the bug’s centripetal acceleration 1.5 s after the bug starts from rest. (1 in = 2.54 cm).

1 answer:

viva [34]3 years ago

3 0

Answer:

6.41m/s2

Explanation:

unit conversion:

78 rev/min = 78rev/min * 2π rad/rev * 1/60 min/sec = 8.168 rad/s

12 in = 12 in * 2.54cm/in = 30.48 cm = 0.3048 m

So the record accelerate from rest to 8.168 rad/s, its (assuming constant) angular acceleration would be

$\alpha = \Delta \omega / t = (8.168 - 0) / 2.67 = 3.06 rad/s^2$

From here we can calculate the angular velocity of the bug at t2 = 1.5s

$\omega_2= \alpha*t_2 = 3.06 * 1.5 = 4.59 rad/s$

Its centripetal acceleration would then be

$a = \omega_2^2*R = 4.59^2*0.3048 = 6.41 m/s^2$

You might be interested in

What is electricity?

igor_vitrenko [27]

A form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current.

5 0

3 years ago

You are performing an experiment that requires the highest possible energy density in the interior of a very long solenoid. Whic

Alinara [238K]

Answer:

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

Explanation:

As we know that energy density depends on the strength of the magnetic field. The magnetic field strength depends on the no of turns of the solenoid and the current passing through it. The greater the number of turns per unit length, greater the current passing through it, more stronger the magnetic field is. As

B = μ₀nI

n = no of turns

I = current through the wire

So the right options are

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

5 0

3 years ago

Carbon (C) and hydrogen (H) are pure substances. Each is made of only one type of atom, but carbon atoms are different from hydr

eimsori [14]

C is the correct answer.

all substances found on the periodic table are elements by definition. anything that is created using elements, such as methane, carbon dioxide, or water, are all compounds.

5 0

3 years ago

Read 2 more answers

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt

Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is $R_z = 4.4 \Omega$

b

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

c

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

d

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

e

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

6 0

3 years ago

I need hellp with thisss plssss????????!!!!

Leto [7]

Answer:

The graph appears to be in error.

The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5

The work done (F * S) is the area of the rhombus

1/2 * (5 +15) * 5 = 50 J

8 0

2 years ago