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3 years ago

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An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t

rue about this situation?
A.
The athlete isn’t doing any work because he doesn’t move the weight.
B.
The athlete isn’t doing any work because he doesn’t hold the weight long enough.
C.
The athlete is doing work because he prevents the weight from falling downward.
D.
The athlete is doing work because 50 kilograms is a significant load to lift.

1 answer:

postnew [5]3 years ago

3 0

Answer:

A. The athlete isn’t doing any work because he doesn’t move the weight.

Explanation:

We must remember the definition of work, which says that work is equal to the product of mass by the distance displaced. In this case, the athlete only does work when he lifts the weight from the ground to the point where he holds the weight suspended.

So when he's holding the weight, he doesn't do any work.

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Suppose a scientist conducts a series of experiments and the results are so amazing she wants to share them with other experts i

nataly862011 [7]

Answer:

The results have not been through the rigorous process of peer review

Explanation:

When a scientist conducts a study and obtains results, those results ought to be submitted to a reputable journal where the results would go through the rigorous protocol of peer review.

During this process, the reliability of the data presented is ascertained before the results are published for other scientists to see.

If the results are hurriedly published on the internet, many researchers who come in contact with the work may be fed with inaccurate information.

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3 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

4 years ago

In still water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles downstream,

Vladimir79 [104]

<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16

Traveling against the current:
(18 - c)*t = 8

Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16

Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16

(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0

Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c

So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3

(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3

And it's verified.</span>

4 0

3 years ago

A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc

Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at $v_c = 23m/s$ speed is

$s_c = \Delta t v_c = 23\Delta t$

The distance traveled by the motorcycle after Δt (seconds) at $m_m = 23 m/s$ speed and acceleration of a = 8 m/s2 is

$s_m = \Delta t v_m + a\Delta t^2/2$

$s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2$

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

$s_m = s_c + 58$

$23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58$

$4 \Delta t^2 = 58$

$\Delta t^2 = 14.5$

$\Delta t = \sqrt{14.5} = 3.807s$

(b)

$s_m = 23 \Delta t + 4 \Delta t^2$

$s_m = 23*3.807 + 58 = 145.581 m$

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3 years ago

What is the mass of an object that has a weight of 2867 N?

Gnom [1K]

Answer:

F = M a

W = M g equivalent equation to express weight of object of mass M

M = W / g = 2867 N / 9.8 m/s^2 = 292.6 kg

7 0

3 years ago