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ycow [4]
3 years ago
9

If you are in a car at a stop sign and then the driver

Physics
2 answers:
Cerrena [4.2K]3 years ago
7 0
The car is accelerating forward.
SVEN [57.7K]3 years ago
4 0

Answer:

it wont be because the hand brake would be on

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Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
What happens when an atoms losses an electron-
inysia [295]
It becomes positive
8 0
2 years ago
The image of an object formed by plane mirror is​
m_a_m_a [10]
Answer of your question is in this photo

8 0
2 years ago
Do quasars reside within or without side of galaxies?
sveticcg [70]

They almost entirely reside within galaxies because quasars are a subset of blackholes with a large and fast enough accretion disk to generate a beam of interstellar material perpendicular to itself. This typically only occurs in the largest black holes at the center of galaxies (supermassive blackholes) or at least stellar black holes---which still occur within galaxies because the material is necessary to form them.

6 0
3 years ago
Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

8 0
3 years ago
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