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2 years ago

What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds

1 answer:

5 0

1250 J in 5 sec= 250 Joule(s) per second (1250/5 0

250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)

250 Watts output = 250/0.65 efficiency = 384 Watts input

1 Horsepower = 732 Watts

Motors 1 Horsepower and under are made in certain step sizes like

3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.

3/4 Horsepower is 549 Watts

1/2 Horsepower is 366 Watts

so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.

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The speed v of a sound wave traveling in a medium that has bulk modulus b and mass density ρ (mass divided by the volume) is v=b

PilotLPTM [1.2K]

As it is given that Bulk modulus and density related to velocity of sound

$v = \sqrt{\frac{B}{\rho}}$

by rearranging the equation we can say

$B = \rho * v^2$

now we need to find the SI unit of Bulk modulus here

we can find it by plug in the units of density and speed here

$B = \frac{kg}{m^3} * (\frac{m}{s})^2$

so SI unit will be

$B = \frac{kg}{m* s^2}$

SO above is the SI unit of bulk Modulus

3 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an

castortr0y [4]

Answer:

Fx = -9.15 N
Fy = 1.72 N
F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

≈ (-9.15309, 1.71982)

The resultant component forces are ...

Fx = -9.15 N
Fy = 1.72 N

Then the magnitude and direction of the resultant are

F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

F∠γ ≈ 9.31∠-10.6°

4 0

3 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

3 years ago

State the condition for maximum current to be drawn from the cell

larisa86 [58]

Use the equation I=V/R where I is current and V is the voltage plus R is the resistance so when voltage is the highest and resistance is lowest the current is the highest

7 0

3 years ago

What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds​

What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds