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1 year ago

A 68 kg runner exerts a force of 59 N. what is the acceleration of the runner

Physics

2 answers:

LuckyWell [14K]1 year ago

8 0

<span>.87 m/s^2 ,hope this helps!!!!!!</span>

romanna [79]1 year ago

5 0

Answer: 0.87m/s²

Explanation: Recall from Newton's second law

Force = mass*acceleration

So,

Acceleration = Force/mass

Acceleration = 59N/68kg

= 0.868N/kg

Which is also 0.87m/s² in standard unit.

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A planet of mass M and radius R has no atmosphere. The escape velocity at its surface is ve. An object of mass m is at rest a di

zubka84 [21]

Answer:

Explanation:

Expression for escape velocity

ve = $\sqrt{\frac{2GM}{R} }$

ve² R / 2 = GM

M is mass of the planet , R is radius of the planet .

At distance r >> R , potential energy of object

= $\frac{-GMm}{r}$

Since the object is at rest at that point , kinetic energy will be zero .

Total mechanical energy = $\frac{-GMm}{r}$ + 0 = $\frac{-GMm}{r}$

Putting the value of GM = ve² R / 2

Total mechanical energy = ve² Rm / 2 r

This mechanical energy will be conserved while falling down on the earth due to law of conservation of mechanical energy . So at surface of the earth , total mechanical energy

= ve² Rm / 2 r

8 0

1 year ago

Suppose you have a container filled with iron and sand. You can separate the iron from the sand if you ____________ so this is a

My name is Ann [436]

A) use a magnet; mixture

5 0

1 year ago

Read 2 more answers

Shows the position-versus-time graph of a particle in SHM. Positive direction is the direction to the right.

BaLLatris [955]

A) t = 0 s, 4 s, 8 s

B) t = 2 s, 6 s

C) t = 1 s, 3 s, 5 s, 7 s

Explanation:

The figure is missing: find it in attachment.

A particle is said to be in Simple Harmonic Motion (SHM) when it is acted upon a restoring force proportional to its displacement, and therefore, the acceleration of the particle is directly proportional to its displacement (but in the opposite direction):

$a\propto -x$

Also, the displacement of a particle in SHM can be described by a sinusoidal function, as shown in the figure for the particle in this problem.

For the particle in this problem, from the figure we can see that the displacement is described by a function in the form

$x(t)=A sin(\omega t)$

where

A is the amplitude

$\omega=\frac{2\pi}{T}$ is the angular frequency, where T is the period. From the graph, we see that the particle completes 1 oscillation in 4 seconds, so

T = 4 s

So the angular frequency is

$\omega = \frac{2\pi}{4}=\frac{\pi}{2}rad/s$

$x(t)=Asin(\frac{\pi}{2}t)$

The velocity of a particle in SHM can be found as the derivative of the displacement, so here we find:

$v(t)=x'(t)=A\omega cos(\omega t) = \frac{A\pi}{2}cos(\frac{\pi}{2}t)$

So, the particle is moving to the right at maximum speed when

$cos(\frac{\pi}{2}t)=+1$

(because this is the maximum positive value for the cosine part). Solving for t,

$\frac{\pi}{2}t=0\\t=0 s$

We also have another time in which this occurs, when

$\frac{\pi}{2}t=2\pi\\\rightarrow t=\frac{4\pi}{\pi}=4 s$

And also when

$\frac{\pi}{2}t=4\pi\\\rightarrow t=8s$

In this case, we want to find the time t at which the particle is moving to the left at maximum speed.

This occurs when the cosine part has the maximum negative value, so when

$cos(\frac{\pi}{2}t)=-1$

Which means that this occurs when:

$\frac{\pi}{2}t=\pi\\\rightarrow t=2 s$

Also when

$\frac{\pi}{2}t=3\pi\\\rightarrow t=6 s$

So, the particle has maximum speed moving to the left at 2 s and 6 s.

The particle is instantaneously at rest when the speed is zero, this means when the cosine part is equal to zero:

$cos(\frac{\pi}{2}t)=0$

This occurs when the argument of the cosine is:

$\frac{\pi}{2}t=\frac{\pi}{2}\\\rightarrow t = 1 s$

Also when

$\frac{\pi}{2}t=\frac{3\pi}{2}\\\rightarrow t = 3 s$

And when

$\frac{\pi}{2}t=\frac{5\pi}{2}\\\rightarrow t = 5 s$

And finally when

$\frac{\pi}{2}t=\frac{7\pi}{2}\\\rightarrow t = 7s$

So, the particle is at rest at t = 1 s, 3 s, 5 s, 7 s.

3 0

1 year ago

What is true of an object pulled inward in an electric field?

slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

5 0

2 years ago

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the

exis [7]

Answer:

113.7

Explanation:

maximum distance (s) = 8.9 km

reference intensity (I0) = 1 x 10^{-12} W/m^{2}

power of a juvenile howler monkey (p) = 63 x 10^{-6} W

distance (r) = 210 m

intensity (I) = power/area

where we assume the area of a sphere due to the uniformity of the output in all directions

area = 4π $r^{2}$ = 4π x $210^{2}$ = 554,176.9 m^{2}

intensity (I) = $\frac{63 x 10^{-6} }{554,176.9} = 113.7 x 10^{-12}$

therefore the desired ratio I/I0 = $\frac{113.7 x 10^{-12}}{1 x 10^{-12}}$ = 113.7

7 0

2 years ago