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Lesechka [4]
3 years ago
12

What type of Galaxy is considered the most common?

Physics
2 answers:
Semmy [17]3 years ago
4 0
Spiral galaxies are most common
lubasha [3.4K]3 years ago
4 0

Answer:

Spiral Galaxies

Explanation:

The most common type galaxy found throughout the universe is the spiral galaxy!

Happy to help. Lol! Good luck!

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A race car accelerates from rest to a velocity of +90 m/s over a distance of 423m. Determine the acceleration of the race car.
dlinn [17]

Answer:

9.57m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity  = 90m/s

Distance  = 423m

Unknown:

Acceleration of the race car  = ?

Solution:

To solve this problem, we should apply one of the appropriate motion equations;

      V²  = U²   + 2as

Where V is the final velocity

           U is the initial velocity

           a is the acceleration

           s is the distance

  90²  = 0²  + 2 x a x 423

  8100 = 846a

      a  = 9.57m/s²

4 0
4 years ago
Look at the diagram. The electricity supplier charges 14p per unit. How much did this electricity cost over the 24 hour period
timama [110]

Total units

  • 31487-31463
  • 23units

Cost°

  • 14(23)
  • $322
4 0
2 years ago
A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T
maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: E_k=\frac{1}{2}mv^2

2. gravitational potential energy of mass m, grav. acc. g and height h: E_g=mgh

3. potential energy in a spring with spring constant k and displacement from equilibrium x: E_s=\frac{1}{2}kx^2

Calculating x:

\frac{1}{2}mv_a^2=\frac{1}{2}kx^2

x=\sqrt{\frac{m}{k}}v_a

Calculating the speed:

\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}

h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R

\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R

Solving for v_b:

v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}

7 0
3 years ago
You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Ap
Inga [223]
The answer is 35 minutes

The Newton's law of cooling is:
T(x) = Ta + (To - Ta)e⁻ⁿˣ

T(x) - the temperature of the coffee at time x
Ta - the ambient temperature
To - the initial temperature
n - constant

step 1. Calculate constant k:

We have:
T(x) = 200°F
x = 10 min
Ta = 68°F
To = 210°F
n = ?

T(x) = Ta + (To - Ta)e⁻ⁿˣ
200 = 68 + (210 - 68)e⁻ⁿ*¹⁰
200 = 68 + 142 * e⁻¹⁰ⁿ
200 - 68 = 142 * e⁻¹⁰ⁿ
132 = 142 * e⁻¹⁰ⁿ
e⁻¹⁰ⁿ = 132/142
e⁻¹⁰ⁿ = 0.93

Logarithm both sides with natural logarithm:
ln(e⁻¹⁰ⁿ) = ln(0.93)
-10n * ln(e) = -0.07
-10n * 1 = - 0.07
-10n = -0.07
n = -0.07 / - 10
n = 0.007

Step 2. Calculate time x when T(x) = 180°F:
We have:
T(x) = 180°F
x = ?
Ta = 68°F
To = 210°F
n = 0.007

T(x) = Ta + (To - Ta)e⁻ⁿˣ
180 = 68 + (210 - 68)e⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79

Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷*ˣ) = ln(0.79)
-0.007x * ln(e) = -0.24
-0.007x * 1 = -0.24
-0.007x = -0.24
x = -0.24 / -0.007
x ≈ 35
5 0
3 years ago
Read 2 more answers
When is the zebra moving and when is it not moving
maw [93]
Its breathing that is moving it is digesting and
6 0
4 years ago
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