In the formula Q = m × c × ΔT, which symbol represents specific heat?

To give an idea of sensitivity of the platypus's electric sense, how far from a 80nC n C point charge does the field have this m

SVETLANKA909090 [29]

The question is incomplete. The complete question is :

A platypus foraging for prey can detect an electric field as small as 0.002 N/C.

-To give an idea of sensitivity of the platypus's electric sense, how far from a +80nC point charge does the field have this magnitude?

Solution :

Given electric field, E = 0.002 N/C

Charge, Q = + 80 nC

$\therefore E = \frac{kQ}{R^2}$

or $R^2=\frac{kQ}{E}$

$R^2=\frac{9\times 10^9 \times 80 \times 10^{-9}}{0.002}$

R = 600 m

This is the distance of the charge from the point of observations.

5 0

3 years ago

The 4.0 -kg head of an ax is moving at 3 m/s when it strikes a log and penetrates 0.01m into the log. What is the average force

Elina [12.6K]

Answer:

the average force the blade exerts on the log is 1791.05 N.

Explanation:

Given;

mass of the ax head, m = 4 kg

speed of the ax, v = 3 m/s

depth traveled into the log, d = 0.01 m

The time to traveled through the depth;

$s = (\frac{u+v}{2} )t\\\\0.01 = (\frac{0+3}{2} )t\\\\0.01 = 1.5t\\\\t = \frac{0.01}{1.5} \\\\t = 0.0067 \ s$

The average force the blade exerts on the log;

$F = ma=\frac{mv}{t} = \frac{4 \times 3}{0.0067} = 1791.05 \ N \\\\$

Therefore, the average force the blade exerts on the log is 1791.05 N.

5 0

3 years ago

Please help!!!!

Dafna11 [192]

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$q=3\cdot 10^{-9}C$ is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

$E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?

ryzh [129]

Answer:

10 N

Explanation:

$f = ma$

= 10m/s^2 * 1 kg

=10N

7 0

3 years ago

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t

Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0

3 years ago