Someone help me please

Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp

erma4kov [3.2K]

Answer:

c. 981 watts

$P=981\ W$

Explanation:

Given:

horizontal speed of treadmill, $v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}$

weight carried, $w=588.6\ N$

grade of the treadmill, $G\%=10\%$

<u>Now the power can be given by:</u>

$P=v.w$

$P=588.6\times\frac{5}{3}$ (where grade is the rise of the front edge per 100 m of the horizontal length)

$P=981\ W$

6 0

4 years ago

In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot

Fittoniya [83]

Answer:

$Q=7.9\times 10^{-10}\ C$

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

$A=5\times 10^{-3}\ m^2$

We Capacitance given as

For air

$C_1=\dfrac{\varepsilon _oA}{d}$

$C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}$

$C_1=2.2\times 10^{-11}\ F$

$C_2=\dfrac{K\varepsilon _oA}{d}$

$C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F$

$C_2=6.6\times 10^{-11}\ F$

Net capacitance

C=C₁+C₂

$C=8.8\times 10^{-11}\ F$

We know that charge Q given as

Q= C V

$Q=12\times 6.6\times 10^{-11}\ C$

$Q=7.9\times 10^{-10}\ C$

6 0

3 years ago

Suppose a 1500 kg speed boat is traveling with a velocity of 30 m/s what is the speed boats Monument

DanielleElmas [232]

Answer: 45000kgm/s

Explanation:

Given that:

Mass of speed boat = 1500 kg

Velocity of speedboat = 30 m/s

Speed boats Momentum = ?

The Speed boat momentum is the product of its mass and the velocity by which it moves. Momentum is a vector quantity and measured in kgm/s

i.e Momentum = mass x velocity

= 1500 kg x 30 m/s

= 45000kgm/s

Thus, the speed boats momentum is 45000kgm/s

7 0

3 years ago

Calculate the equivalent resistance for both circuits. Series circuit: 2 Ω and 4 Ω Parallel circuit: 2 Ω and 4 Ω Which circuit h

goldenfox [79]

Equivalent resistance is also known as the overall resistance.

For resistors in a series circuit, the total resistance is computed using the formula:

$R_{T} = R_{1}+ R_{2}+ R_{3}... R_{n}$

In other words, you just add up the resistance of each resistor in the series circuit. In your case you only have two resistors. You have 2Ω and 4Ω. So all you need to do is add that up.

$R_{T} = R_{1}+ R_{2}$
$R_{T} = 2 + 4=6$

The total resistance of the series circuit is 6Ω

In a parallel circuit you get the total resistance using the formula:
$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}$

First you get the sum of all fractions and at the end take the reciprocal of the resulting fraction and divide. So let us take your problem into consideration where you have two resistors that have a resistance of 2Ω and 4Ω.

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}$
$\frac{1}{R_{T}} = \frac{1}{2}+\frac{1}{4}$
$\frac{1}{R_{T}} = \frac{2}{4}+\frac{1}{4}$
$\frac{1}{R_{T}} = \frac{3}{4}$

Get the reciprocal of the resulting fraction 3/4 and then divide. The reciprocal of 3/4 is 4/3.

4/3 = 1. 33Ω

So if you compare the equivalent resistance of the two circuits, the series circuit has a higher equivalent resistance.

3 0

4 years ago

Read 2 more answers

Electric power companies measure energy consumption in kilowatt-hours, denoted kWh. One kilowatt-hour is the amount of energy tr

Natali [406]

The amount of energy used in the billing period is 5,400,000,000 joules.

One kWH is the amount of energy transferred in one hour, there 1 kWh is equal to

1 kWh=1*1000*(J/s)*3600 s

=3600000 J

Thus the amount of energy in joules consumed by the user for the billing period is =1500*3600000=5,400,000,000 J.

The amount of energy used in the billing period is 5,400,000,000 joules.

7 0

4 years ago

Someone help me please ​

Someone help me please