All categories

2 years ago

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics

2 answers:

almond37 [142]2 years ago

8 0

That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.

yawa3891 [41]2 years ago

6 0

The red at the bottom is just the holder for the beaker. It’s not apart of the density.

You might be interested in

1. Straight-in spaces can leave you a safer out if

Juliette [100K]

Answer:

C. you're able to reverse out of the parking spot

Explanation:

Straight-in parking is an approach of parking that allows a more flexible traffic layout where a driver can approach the spot from either direction and still safely park within the lines. It thus helps to prevent blockage of cars. Each car can move in and out freely preventing it from congestion.

This way of parking can leave you safe when you able to reverse out of the parking spot. It gives you greater control and makes it easier to maneuver out space. The benefits of Straight-in parking are,

- Allows for two-way traffic

- Drivers can line up the vehicle from multiple angles

- Saves time for drivers

3 0

3 years ago

Two objects attract each other gravitationally. If the distance between their centers decreases by a factor of 2, how does the g

kramer

Answer:

The gravitational force between them increases by a factor of 4

Explanation:

Gravitational force is a force of attraction between two objects with masses M and m which are separated by a distance R. It is given mathematically as:

Fg = GMm/R²

Where G = Gravitational constant.

If the distance between their centers, R, decreases by a factor of 2, then it means the new distance between their centers is:

r = R/2

Hence,the gravitational force becomes:

Fg = GMm/r²

Fg = GMm/(R/2)²

Fg = GMm/(R²/4)

Fg = 4GMm/R²

Hence,the gravitational force increases by a factor of 4.

6 0

3 years ago

At what point in its motion is the KE of a pendulum bob a maximum? 1. The KE does not change. 2. at the lowest point 3. at the h

Andrew [12]

Answer:

2. at the lowest point

Explanation:

The motion of the pendulum is a continuous conversion between kinetic energy (KE) and gravitational potential energy (GPE). This is because the mechanical energy of the pendulum, which is sum of KE and GPE, is constant:

E = KE + GPE = const.

Therefore, when KE is maximum, GPE is minimum, and viceversa.

So, the point of the motion where the KE is maximum is where the GPE is minimum: and since the GPE is directly proportional to the heigth of the bob:

$GPE=mgh$

we see that GPE is minimum when the bob is at the lowest point,so the correct answer is

2. at the lowest point

3 0

3 years ago

Organizing and interpreting information from the senses is called

PIT_PIT [208]

I believe the answer you are looking for is perception.

3 0

3 years ago

Read 2 more answers

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

2 years ago