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Arada [10]
3 years ago
6

Ben and Jerry, arch rivals, decide to have a weight lifting contest during PE. Ready, set, go! Ben and Jerry both lift a 250 kg

barbell 10 times over their heads. They are approximately the same height and lift the barbell the same distance in the air. It takes Ben 5 seconds to complete 10 lifts; it takes Jerry 25 seconds to complete his 10 lifts. Which statement is MOST accurate regarding the weightlifting contest? A) Ben did more work than Jerry. B) Ben has more power than Jerry. C) Ben and Jerry have the same power. D) Ben does more work and is more powerful than Jerry.
Physics
2 answers:
Sidana [21]3 years ago
7 0

The answer is B. Your welcome.

B    ) STILL CHILL

cluponka [151]3 years ago
7 0

Answer:

B) Ben has more power than Jerry.

Explanation:

Here work done to lift the mass to the certain height is given as

W = mgh

now given that Ben and Jerry both lift the mass to the same height 10 times

so here work done by Ben and Jerry must be same.

now we know that power is defined as rate of work done

P = \frac{W}{t}

now we know from above formula that it took less time by Ben as he lift the mass 10 times in 5 seconds while Jerry took 25 seconds to lift the mass 10 times.

So here power by Ben will be more than Jerry as is took lesser time to complete the task of 10 times lift of mass

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One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista
devlian [24]

Answer:

2.24 m/s²

Explanation:

Using equation of motion

s = ut + \frac{1}{2}at²

u = 0 , t = 3.17 s , s = 11.26 m

Put these values in the equation above

11.26 = 0 +.5 x a( 3.17)²

a = 2.24 ms⁻².

So acceleration due to gravity on that planet will be 2.24 m s⁻².

7 0
3 years ago
An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi
denpristay [2]

Answer:

Explanation:

An information contains

25Hz and 75Hz sine wave

Sample frequency is 500Hz

The analogy signal are generally

y(t) = Asin(2πx/λ - wt), w=2πf

y1(t)=Asin(2πx/λ - wt)

y1(t)=Asin(2πx/λ - 2π•25t)

y1(t)=Asin(2πx/λ - 50πt)

Similarly

y2(t)=Asin(2πx/λ - 150πt)

Using Nyquist theorem

Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.

From sampling

f(nyquist)=f(sample)/2

f(nyquist)=500/2

f(nyquist)=250Hz

From signal

The highest frequency is 150Hz

F(nyquist) = 2×F(highest)

f(nyquist)= 2×150

f(nyquist)= 300Hz

Sample per frequency Ns is given as

Ns=F(sample)/F(highest signal)

Ns=500/150

Ns=3.33sample/period

This is above nyquist rate of 2sample/period

So signal below 300Hz reproduced without aliasing.

The highest resulting frequency is 300Hz

6 0
3 years ago
. Inside a conducting sphere of radius 1.2 m, there is a spherical cavity of radius 0.8 m. At the center of the cavity is a poin
MAVERICK [17]

Answer:

 E = 1,873 10³ N / C

Explanation:

For this exercise we can use Gauss's law

        Ф = E. dA = q_{int} / ε₀

Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.

The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.

        The surface of a sphere is

             A = 4π r²

             E 4π r² = q_{int} /ε₀

  The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is

           q_{int} = q₁ + q₂

           q_{int} = (530 - 200) 10⁻⁹

           q_{int} = 330 10⁻⁹ C

The electric field is

             E = 1 / 4πε₀   q_{int} / r²

            k = 1 / 4πε₀

            E = k q_{int}/ r²

Let's calculate

           E = 8.99 10⁹   330 10⁻⁹/ 1.32²

           E = 1,873 10³ N / C

7 0
3 years ago
A block is 10cm long, 5cm wide and 2cm high and weighs 100g. What is the volume of the block? What is the density?
Ivahew [28]

Answer:

1gm/cm^3

Explanation:

its the answer

5 0
3 years ago
In Ørsted’s observation, the current-carrying wire acted like a
hichkok12 [17]
B, C. Also literally a quick search yielded these results, roughly half the time to type this out. 
6 0
3 years ago
Read 2 more answers
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