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cestrela7 [59]
3 years ago
13

Detailed measurements of the disk and central bulge region of our Galaxy suggest our Milky Way is a:Select one:A. quasar. B. bar

red spiral galaxy. C. very flat elliptical galaxy. D. very dusty irregular galaxy. E. normal spiral galaxy.
Physics
1 answer:
chubhunter [2.5K]3 years ago
4 0

I believe the answer is:

~D. Very dusty irregular galaxy.

Hope this helps!!!

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Part D
anygoal [31]

Answer: I didn't see a difference because the large ball's vertical displacement and velocity are the same as the small one's.

Explanation:

5 0
3 years ago
A research submarine has a 20-cm-diameter window that is 9.0 cm thick. The manufacturer says the window can withstand forces up
Ratling [72]

Answer:

The maximum safe depth in salt water is 3758.2 m.

Explanation:

Given that,

Diameter = 20 cm

Radius = 10 cm

Thickness = 9.0 cm

Force F = 1.2\times10^{6}\ N

Inside pressure = 1.0 atm

We need to calculate the area

Using formula of area

A=\pi\times r^2

Put the value into the formula

A=\pi\times(10\times10^{-2})^2

A=0.0314\ m^2

We need to calculate the pressure

Using formula of pressure

P=\dfrac{F}{A}

Put the value into the formula

P=\dfrac{1.2\times10^{6}}{0.0314}

P=38216560.50\ Pa

P=3.8\times10^{7}\ Pa

We need to calculate the maximum depth

Using equation of pressure

P=P_{atm}+\rho gh

h=\dfrac{P-P_{atm}}{\rho g}

Put the value into the formula

h=\dfrac{3.8\times10^{7}-101325}{1029\times9.8}

h=3758.2\ m

Hence, The maximum safe depth in salt water is 3758.2 m.

4 0
3 years ago
2 When a cube of hot metal is placed in a beaker of cold water, the temperature of the water -
jek_recluse [69]
The answer to this is B
4 0
3 years ago
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0
3 years ago
1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)
cupoosta [38]

Answer:

1) k = 10 [N/m]

2) a-) x = 0.4 [m]

b)  x = 0.075 [m]

Explanation:

To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.

F = k*x

where:

F = force [N] (units of Newtons]

k = spring constant  [N/m]

x = distance = 10 [cm] = 0.1 [m]

Now, the weight is equal to the product of the mass by the gravity

W = m*g = F

where:

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration = 10 [m/s²]

F = 0.1*10 = 1 [N]

Now clearing k

k = F/x

k = 1/0.1

k = 10 [N/m]

2)

a ) if the force is 4 [N]

clearing x

x = F/k

x = 4/10

x = 0.4 [m]

m = 75 [g] = 0.075 [kg]

W = m*g = F

F = 0.075*10 = 0.75 [N]

x = .75/10

x = 0.075 [m]

5 0
2 years ago
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