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pochemuha
3 years ago
10

What are the three ways to accelerate?

Physics
2 answers:
Nookie1986 [14]3 years ago
5 0

Answer:

There are three ways an object can accelerate: a change in velocity, a change in direction, or a change in both velocity and direction.

Explanation:

Ivanshal [37]3 years ago
5 0
They’re rigth!! ^^ they gave u the answer already
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if the earth were to cease rotating about its axis,then what is the increase in the value of gravity?
IrinaK [193]
There is no adjustment in gravity, yet there is an adjustment in 'weightness'. 
Gravitational compel and weight with respect to an edge are not similar things, despite the fact that it is normally educated something else. 
Weight is really the aggregate of gravitational powers and of inertial drive for a question very still (no Coriolis compel) in a given casing. 
In the event that the Earth were not pivoting, weight would increment most at the Equator and be unaltered at the Poles.
3 0
3 years ago
A local fun house incorporates a gently curved, concave, spherical mirror into its display. When a child stands 1.2 m from the m
lys-0071 [83]

Options:

a) more than 0.8 m .

b) equal to 0.8 m .

c) between 0.5 m and 0.8 m .

d) less than 0.5 m .

Answer:

b) equal to 0.8 m .

Explanation:

Note:

An upside down image = Inverted Image

An image that appears in front of the mirror = Real image

An image that appears behind the mirror = Virtual image

Let the object distance from the pole of the mirror be u

When the child stands 1.2 m from the mirror:

u = 1.2 m ( Real and Inverted image of the child is formed)

When the child stands about 0.8 m from the mirror:

u = 0.8 m (Virtual, erect and magnified image of the child is formed)

When the child stands about 0.5 m from the mirror:

u = 0.5 m ( Virtual and erect image of the child is formed)

Note: All objects positioned behind the focal length of a concave mirror are always real. Objects start becoming virtual when they are placed on the focal length or in front of it (Close to the pole of the mirror), although objects placed on the focus has its image formed at infinity.

Since the nature of the image formed changed from real to virtual when the child stands about 0.8 m from the mirror, then the focal length is approximately equal to 0.8 m

3 0
3 years ago
The Schwarzschild radius RBH for an object of mass M is defined as: Rbh = (2GM)/(c^2)
Agata [3.3K]

Answers:

1) 2951.39 m

2) 0.000952 m

3) 8.765(10)^{16}kg

Explanation:

1) We know the Schwarzschild radius RBH is given by the following equation:

RBH=\frac{2GM}{c^{2}}   (1)

Where:

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Universal Gravitational Constant

M the mass of the black hole

c=3(10)^{8}m/s is the speed of light

Now, if we have a black hole with the mass of the Sun (M_{Sun}=1.99(10)^{30}kg), its radius will be:

RBH_{Sun}=\frac{2GM_{Sun}}{c^{2}}   (2)

RBH_{Sun}=\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1.99(10)^{30}kg)}{{(3(10)^{8}m/s)}^{2}}   (3)

RBH_{Sun}=2951.39m   (4) This is the radius of the black hole with the mass of the Sun

2) On the other hand, if the black hole has the mass of Mars (M_{Mars}=6.42(10)^{23}kg), its radius will be:

RBH_{Mars}=\frac{2GM_{Mars}}{c^{2}}   (5)

RBH_{Mars}=\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.42(10)^{23}kg)}{{(3(10)^{8}m/s)}^{2}}   (6)

RBH_{Mars}=0.000952m   (7) This is the radius of the black hole with the mass of Mars

3) In this case, we have to isolate M from (1):

M=\frac{RBH c^{2}}{2G}   (8)

Where RBH=1.30(10)^{-10}m

Solving (8) with the known values:

M=\frac{(1.30(10)^{-10}m)(3(10)^{8}m/s)^{2}}{2(6.674(10)^{-11}m^{3}/kgs^{2}}   (9)

M=8.765(10)^{16}kg   (10) This is the mass of the black hole

7 0
3 years ago
What happen when a light from the sun passes through any type of matter
mixer [17]
It creates a shadow!
5 0
3 years ago
A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. When the temperature of th
Nesterboy [21]

Answer:

Option C is the correct answer.

Explanation:

By Charles's law we have

        V ∝ T

That is

       \frac{V_1}{T_1}=\frac{V_2}{T_2}

Here given that

      V₁ = 0.20 cubic meter

      T₁ = 333 K

      T₂ = 533 K

Substituting

      \frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\\frac{0.20}{333}=\frac{V_2}{533}\\\\V_2=\frac{0.20}{333}\times 533=0.3198m^3

New volume of the gas  = 0.3198 m³

Option C is the correct answer.

7 0
4 years ago
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