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3 years ago

Help please this is worth 30% of my grade in total

Physics

1 answer:

Dmitrij [34]3 years ago

6 0

Answer:

D. Both are held together by gravitational forces

Explanation:

The common features in galaxies are;

Gravity holds the stars together
The dense region in a galaxy is the center
They all have a faint outer region and mysterious dark matter halo
They all have stars, planets, asteroids, galactic cloud

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A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they h

Lilit [14]

Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)

In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:

132.33km/h - 200km/h = -67.69km/h

km and the position in "y" is equal to 137km*sin(15°) = 35.4km

This means that the velocity of the air in the y-axis is 35.4km/h

So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

8 0

3 years ago

Please help meeeee!!!

Sati [7]

Answer:

that answer is

b. 1/4,000

8 0

3 years ago

Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it

olga2289 [7]

<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>

6 0

4 years ago

PLS HELP FOR THIS PHYSICS TEST <br><br> (pls only answer if u rlly know, this is an important test)

lesya692 [45]

Answer:

C. 8.01 m/s²

Explanation:

vf²= vi² + 2 • a • d

2ad = vf² - vi²

a = (vf²- vi²)/2d

d=25.00 -5.00=20.00 m

vi =0

vf=17.90 m/s

a =(17.90² -0²)/(2*20) = 8.01 m/s²

4 0

3 years ago

A calorimeter is a container that is insulated from the outside, so a negligible amount of energy enters or leaves the container

Len [333]

Answer:

This is a heat balance question with negligible heat losses to the surroundings.

Sum of heat losses = Sum of the heat gains

We are required to find the final temperature of the calorimeter system T1?

we were given

C = specific heat capacities of Copper c, Water w, Ice i and Lead l

m = masses as mc = 0.1kg, mw = 0.16kg, mi=0.018kg, ml=0.75kg,

T = Temperature of the components, Lead Tl= 225C and at ambient pressure, Temp. To of the water, ice and copper = 25C

Since, at thermal equilibrium after adding the Lead, the lead temperature Tl decreases while copper, water and ice temp decrease, so we have respective heat losses 'q' as,

qc + qw + qi + ql =0

which means that

(mc x Cc x ΔTC) + (mw x Cw x ΔTC) + (mi x Ci x ΔTC) + (ml x Cl x ΔTC) = 0

(0.1x390x (T1 - 25)) + (0.16x4190x (T1 - 25)) + (0.018x4190x (T1 - 0)) + (0.750x130x (T1 - 225)) = 0

882.32T1 - 41558 = 0

T1 = 47.1K

Hence, the final temperature of the calorimeter system is 47.1K

3 0

4 years ago