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3 years ago

Can someone help me please??

Physics

1 answer:

Mrrafil [7]3 years ago

7 0

1. Our solar system is the only place in the universe where gravity played a key part in the formation of planets.

2. Rocky planets are small, dense, and orbit relatively close to the sun, compared to the Jovian planets, which are large, less dense, and orbiting far from the sun.

3. _______

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What material composes most of the volume of planets jupiter and saturn?

wlad13 [49]

I think the answer would be

6 0

3 years ago

Which objects would be considered to be in free fall?

SVEN [57.7K]

Answer:

Any object that is being acted upon only by the force of gravity is said to be in a state of free fall.

Explanation:

8 0

3 years ago

Read 2 more answers

Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of

djverab [1.8K]

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

$F_c=ma_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

$F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN$

3 0

3 years ago

A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po

VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

5 0

3 years ago