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Deffense [45]
3 years ago
12

Astronomers are comparing two stars that are known to have the

Physics
1 answer:
timurjin [86]3 years ago
6 0
It is 4 times closer
You might be interested in
Describe the initial horizontal and vertical velocity of a horizontally launched projectile on Earth, as well as what happens to
Jobisdone [24]

Answer:

Explained below

Explanation:

To explain this, let's consider a tennis ball being launched from the top of a very high building.

Now, if the tennis ball is launched horizontally without any upward angle but with an initial velocity of 10 m/s. In this motion, If there is no gravity, the tennis ball would continue in motion at that same speed of 10 m/s in the horizontal direction. However, in reality, gravity causes the tennis ball to accelerate downwards at a rate of 9.8 m/s for every second. This implies that the vertical velocity component is changing at the rate of 9.8 m/s every second.

Thus, after 1 second, horizontal velocity component will remain 10 m/s and vertical component will be 9.8 m/s × 1 = 9.8 m/s downwards.

Also, after 2 seconds, the vertical velocity component will remain 10 m/s, however the vertical component will now be 9.8 × 2 = 19.6 m/s downwards.

Same procedure is repeated as t increases by 1 second.

5 0
2 years ago
A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an
Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

<u>ΔP = 0.056 psi</u>

8 0
3 years ago
To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be
Troyanec [42]
Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement 
so when we wanted to move a 100kg a distance of 1m 
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m 

by the same way:-
------------------------
work in = 100kg * 2m * g (m/s^2)= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m
6 0
2 years ago
Read 2 more answers
A SMA wire in the un-stretched condition is then given an initial strain of εo (to preload the wire) at room temperature (RT). I
julia-pushkina [17]

Answer: tensional force

Explanation:

Tension force on a material occurs when two equal forces act on a material in an opposite direction away from the ends of the material.

Pre-tensing a wire material increases its load bearing capacity and reduces its flexure.

6 0
3 years ago
As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
fgiga [73]

Answer: 107.8\ J

Explanation:

Given

Initial position of object is (4.4 i+5 j)

Final position of object is (11.6 i -2 j)

Force acting (4i-9j)

Work done is given by

\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J

Initial kinetic energy

K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J

Change in kinetic energy is equal to work done by object

\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ J

5 0
3 years ago
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