Calculate the power if the current in a circuit is 3A and the resistance of the circuit is 50hms

When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed sh

aev [14]

Answer:

Explanation:

Given

Initial speed is u=V

Maximum height of Pebble is H

Deriving maximum height of Pebble and considering motion in vertical direction

$v^2-u^2=2 as$

where v=final velocity

u=initial velocity

a=acceleration

s=Displacement

Final velocity will be zero at maximum height

$0-(V)^2=2\times (-g)\cdot H$

$H=\frac{V^2}{2g}$

i.e. maximum height is dependent on square of initial velocity

for twice the height

$2H=\frac{(V')^2}{2g}$

on comparing

$V'=\sqrt{2}V$

7 0

3 years ago

The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot

KiRa [710]

Answer:

$T_{f}$ = 85.7 ° C

Explanation:

For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state

Q₁ = m L

Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water

Q₁ = 2.00 10⁻³ 2.26 10⁶

Q1 = 4.52 10³ J

Now the heat of coffee in the cup, which does not change state is

Q coffee = M $c_{e}$ ( $T_{f}$ - $T_{i}$ )

Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat

Qc = - Q₁

M ce ( $T_{f}$ - $T_{i}$ ) = - Q₁

The coffee dough left in the cup after evaporation is

M = 250 -2 = 248 g = 0.248 kg

$T_{f}$ -Ti = -Q1 / M $c_{e}$

$T_{f}$ = Ti - Q1 / M $c_{e}$

Since coffee is essentially water, let's use the specific heat of water,

$c_{e}$ = 4186 J / kg ºC

Let's calculate

$T_{f}$ = 90.0 - 4.52 103 / (0.248 4.186 103)

$T_{f}$ = 90- 4.35

$T_{f}$ = 85.65 ° C

$T_{f}$ = 85.7 ° C

5 0

3 years ago

A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At

STALIN [3.7K]

Answer:

(a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

Explanation:

Given that,

Nozzle diameter = 0.25 in = 0.00635 m

Hose pipe diameter = 0.64 in = 0.016256 m

Pressure = 1.9 atm =192518 Pa

(a). We need to calculate the speed of the water in the nozzle

Flow Speed at the inlet pipe will be given by using Continuity Equation

$Q_{1}=Q_{2}$

$v_{1}A_{1}=v_{2}A_{2}$

$v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})$

Where, A = area of pipe

$A=\pi\times \dfrac{d^2}{4}$

$v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})$

Put the value into the formula

$v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}$

$v_{1}=3.014\ m/s$

The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}$

Where, $h_{1}=h_{2}$

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2$

$P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)$

Put the value into the formula

$P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)$

$P_{1}=188081.702\ Pa$

$P=1.86\ atm$

Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

7 0

3 years ago

Friction is necessary when you are on a bike to stay

zzz [600]

Answer:

yes friction is needed hope this helps might of been to long tho

7 0

3 years ago

Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45

victus00 [196]

<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>

4 0

3 years ago