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kkurt [141]
3 years ago
6

What color does the star alpha centauri appear to be?

Physics
2 answers:
myrzilka [38]3 years ago
3 0
B.yellow-white is the answer.
vladimir2022 [97]3 years ago
3 0

I did the test and I got b yellow-white.

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A small airplane takes on 302 l of fuel. If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?
ella [17]

To calculate the mass of the fuel, we use the formula

m = V \times  \rho

Here, m is the mass of fuel, V is the volume of the fuel and its value is V =302 \ L =  302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL and  \rho is the density and its value of  0.821 g/mL.

Substituting these values in above relation, we get  

m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg

Thus, the mass of the fuel 247 .94 kg.

8 0
3 years ago
The vehicles are fitted with shockers why​
MissTica

Shockers acts as a shock absorber in the vehicle making the ride comfortable. When vehicle goes on rough road then it gets shocks from the road deformities the springs present in the shockers provide counter force and thus making the vehicle stable ride without shock.

[I copied this from the internet, but hope it helps!]

7 0
3 years ago
Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

g=9.8 m/s^2 is the acceleration due to gravity

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

5 0
3 years ago
When dissolved into water, a molecule of potassium chloride ___________.
avanturin [10]

Answer:

ttrttrv vt vrtttv

Explanation:

4 0
2 years ago
Read 2 more answers
A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was
AlekseyPX

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

(c) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx

Δx = 50 m

7 0
3 years ago
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