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3 years ago

A runner starts from rest and stops in 12 seconds. He covers

Physics

1 answer:

Schach [20]3 years ago

6 0

Answer:

b 1.39 m/s²

Explanation:

Given the following data;

Time = 12 seconds

Distance, S = 100 m

Since it's starting from rest, the initial velocity is equal to 0m/s.

To find the acceleration, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 0(12) + ½*a*12²

100 = 0 + 72

100 = 72a

Acceleration, a = 100/72

Acceleration, a = 1.389 ≈ 1.39 m/s²

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A 0.12 kg bird is flying at a constant speed of 7.8 m/s. what is the birds conetic energy?

lana [24]

KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
KE= 3.7 J - answer
Hope this helps

7 0

3 years ago

Read 2 more answers

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

3 years ago

You are packing for a trip to another star. During the journey, you will be traveling at 0.99c. You are trying to decide whether

Elenna [48]

Answer:

Do neither of these things ( c )

Explanation:

For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula

L = (l) $\sqrt{1 -\frac{v^{2} }{c^{2} } }$

where l = Measured distance from object at rest, L = contracted measured in relation to the observer , v = speed of clock , c = speed of light

you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .

7 0

3 years ago

When a satellite is in orbit around the earth the force of gravity on the satellite-

mojhsa [17]

Answer:

Is always towards the center of the Earth

Explanation:

As a satellite moves around the Earth in a circular orbit, the direction of the force of gravity is always towards the center of the Earth. At an altitude of 100 km, you would be so high that you would see black sky and stars if you looked upwards.

5 0

3 years ago

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

$\theta$ = Angle of slide = 20.7°