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DerKrebs [107]
3 years ago
12

A space shuttle with a mass of approximately 7.08 E5 kg is sitting on the launch pad. What would be the weight of the space shut

tle as it prepares for launch?
If there’s any chance work can be shown that would be great, the equation we were told to work with is Wo=mog
Physics
1 answer:
Afina-wow [57]3 years ago
6 0

The weight of the shuttle is 6.94\cdot 10^6 N

Explanation:

The weight of an object on Earth is the gravitational force exerted by the Earth on the object.

The magnitude of the weight of an object is given by:

W=mg

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity on Earth's surface

And the direction is downward (towards the Earth's centre).

For the shuttle in this problem, its mass is

m=7.08\cdot 10^5 kg

So, its weight is

W=(7.08\cdot 10^5)(9.8)=6.94\cdot 10^6 N

Note that while the mass of an object (m) does not change, its weight (W) changes according to the location, since the value of g can be different at different location (for example, on the Moon, the value of g is about 1/6 the value of g on the Earth).

Learn more about weight and forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc
motikmotik

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

E = mc^2

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

E = mc^2

4 0
3 years ago
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The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v
velikii [3]

Answer:

\theta = n\pi/2, {\rm where~n~is~an~integer.}

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k

Similarly,

\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)

So,

\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)

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7 0
3 years ago
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside thi
Zanzabum

Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

ΔV = EL

where;

ΔV is change in electric potential

E = ΔV / L

E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

7 0
3 years ago
Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per
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Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90^{\circ} to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

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