Please answer correctly will give brainlist if correct

A 0.155 kg arrow is shot upward

Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0

2 years ago

after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

2 years ago

If a FM radio station broadcasts at 80. 3 MHz (megahertz), what is its wavelength in m (speed of light 3. 0 x 108 m/s)

hammer [34]

Answer:

Wavelength = 3.74 m

Explanation:

In order to find wavelength in "metres", we must first convert megahertz to hertz.

1 MHz = 1 × 10⁶ Hz

80.3 Mhz = <em>x</em>

<em>x </em>= 80.3 × 1 × 10⁶ = 8.03 × 10⁷ Hz

The formula between wave speed, frequency and wavelength is:

v = fλ [where v is wave speed, f is frequency and λ is wavelength]

Reorganise the equation and make λ the subject.

λ = v ÷ f

λ = (3 × 10⁸) ÷ (8.03 × 10⁷)

λ = 3.74 m [rounded to 3 significant figures]

8 0

2 years ago

A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur

Anna35 [415]

Answer:

$\dot{W_{H} } = 4244.48 Btu/h$

Explanation:

Temperature of the house, $T_{H} = 70^{0} F$

Convert to rankine, $T_{H} = 70^{0}+ 460 = 530 R$

Heat is extracted at 40°F i.e $T_{L} = 40^{0}F = 40 + 460 = 500 R$

Calculate the coefficient of performance of the heat pump, COP

$COP = \frac{T_{H} }{T_{H} - T_{L} } \\COP = \frac{530 }{530 - 500 }\\ COP = \frac{530}{30} \\COP = 17.67$

The minimum power required to run the heat pump is given by the formula:

$\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\$ ...............(*)

Where the heat losses from the house, $\dot{Q_{H} } = 75,000 Btu/h$

Substituting these values into * above

$\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h$

3 0

2 years ago

Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

$T=575.16K$

4 0

3 years ago