According to the Periodic Table of Elements, Calcium (Ca) has:

An oscillator creates periodic waves on a stretched string.

sattari [20]

Answer:

A. The wavelength doubles but the wave speed is unchanged

Explanation:

The relationship between the period and wavelength is direct. Doubling the period of the oscillator will correspondingly double the wavelength but the wave speed is unaffected

6 0

3 years ago

An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh

never [62]

Answer:

(a) 46.94 J.

(b) 55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0

2 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

2 years ago

In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el

kkurt [141]

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 ) x 25 = 2083.33 mV = 2.083 V.

7 0

3 years ago

A baseball pitcher throws a 0.14 kg ball toward a batter who is 18 m away.

Olin [163]

Answer:

A. 112 J

Explanation:

KE = ½mv² = ½(0.14)40² = 112 J

5 0

3 years ago