What is a key phrase for remembering the order of the planets?

In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c

Bess [88]

Answer:

Part a)

$dF = -\frac{mv^2}{r^2} dr$

Part b)

$dF = \frac{2mvdv}{r}$

Part c)

$dT = - \frac{2\pi r}{v^2} dv$

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

$F = \frac{mv^2}{r}$

now variation in force is given as

$dF = -\frac{mv^2}{r^2} dr$

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

$F = \frac{mv^2}{r}$

so we have

$dF = \frac{2mvdv}{r}$

Part c)

As we know that time period of the circular motion is given as

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

so here if radius is constant then variation in time period is given as

$dT = - \frac{2\pi r}{v^2} dv$

8 0

3 years ago

Waves on a pond are an example of which kind of wave?

KATRIN_1 [288]

<span>
</span><span>Waves on a pond are an example of which kind of wave?

</span>B. surface waves

6 0

4 years ago

a horse runs with an initial velocity of 11m/s and slows to 5.2 m/s over a time interval of 3.1 s what is the horse's average ac

Katarina [22]

Answer:

a = change in v / change in time

= (5.2 - 11) / 3.1

= -1.87 m/s^2

Explanation:

6 0

3 years ago

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu

Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰, force = 4.07 N

(b) When the angle, θ = 90 ⁰, force = 4.7 N

(c) When the angle, θ = 120 ⁰, force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

$F = BIl \ sin(\theta)$

(a) When the angle, θ = 60 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N$

(b) When the angle, θ = 90 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N$

(c) When the angle, θ = 120 ⁰

$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N$

6 0

3 years ago

The 6 strings on a guitar all have about the same length and are stretched with about the same tension.The highest string vibrat

ahrayia [7]

Answer:

B. d(low)=4d(high)

Explanation:

Frequency of a string can be written as;

f = v/2L

Where;

v = sound velocity

L = string length

Frequency can be further expanded to;

f = v/2L = (1/2L)√(T/u) ......1

Where;

m= mass,

u = linear density of string,

T = tension

p = density of string material

A = cross sectional area of string

d = string diameter

u = m/L .......2

m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)

f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))

f = (1/2L)√(T/((p(πd^2)L/4)/L))

f = (1/2L)√(4T/pπd^2)

f = (1/L)(1/d)√(4T/pπ)

Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.

f ~ 1/d

So, if

4f(low) = f(high)

Then,

d(low) = 4d(high)

6 0

3 years ago