The woman in the first photo lifts a plant 2 m

b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 2.50 m/s

fomenos

Answer:

The horizontal distance is 0.64 m.

Explanation:

Initial velocity, u =2.5m/s

The maximum horizontal distance is

$R = \frac{u^2}{2g}\\\\R = \frac{2.5\times 2.5}{9.8}\\\\R = 0.64 m$

3 0

2 years ago

What is the resistance of the coil A at 600 kelvin if its resistance at 300 kelvin is 50 ohms? (Assume the temperature coefficie

leonid [27]

155Ω

Explanation:

R = R ref ( 1 + ∝ ( T - Tref)

where R = conduction resistance at temperature T

R ref = conductor resistance at reference temperature

∝ = temperature coefficient of resistance for conductor

T = conduction temperature in degrees Celsius

T ref = reference temperature that ∝ is specified at for the conductor material

T = 600 k - 273 k = 327 °C

Tref = 300 - 273 K = 27 °C

R = 50 Ω ( 1 + 0.007 ( 327 - 27) )

R = 155Ω

8 0

3 years ago

Read 2 more answers

All of the orbitals in a given subshell have the same value of the __________ quantum number.

Gelneren [198K]

All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number

Hope this helps!</span>

3 0

2 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

or

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

or

$f=\frac{33.84}{0.935}$

or

$f=36.19\approx 36\ \textup{Hz}$

5 0

2 years ago

Using your best estimates, how many times would you have to slap a 1 kg rotisserie chicken in order to cook it? You can assume t

FromTheMoon [43]

Answer:

n= 16021.03 slaps

Explanation:

Using law of Energy conservation

E_{thermal}= Kinetic energy of hand

⇒ $mc\Delta T= n\frac{1}{2}m_hv_h^2$

m_h= mass of the hand = 0.4 kg

v_h= velocity of the hand = 10 m/s

n= number of slaps

c= 4180 J/Kg °C

m= mass of chicken = 1 kg

Assuming all the energy of hand goes into chicken

Given Ti=0°C and T_f= 170 F= 76.66°C

Now putting the values in above equation to get n

$1\times4180(76.66)= n\frac{1}{2}0.4\times10^2$

n= 16021.03 slaps

8 0

2 years ago