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Arturiano [62]
3 years ago
8

What force attracts an object to earth. help

Physics
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

Gravity. the weight force or weight of an object is the force because of gravity which acts on the object attracting it towards the center of the earth mass is a measure of the amount of material or matter that an object contains

You might be interested in
The expression below was formed by combining different gas laws.
kap26 [50]

Answer:

Avogadro's law.

Explanation:

Avogadro’s law states that, equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

Mathematically,

V n

V = Kn where V = volume in cm3, dm3, ml or L; n = number of moles of gas;

K = mathematical constant.

The ideal gas equation is a combination of Boyle's law, Charles' law and Avogadro’s law.

V 1/P at constant temperature (Boyle’s law)

V T at constant pressure ( Charles’law)

V n at constant temperature and pressure ( Avogadro’s law )

Combining the equations yields,

V nT/P

Introducing a constant,

V = nRT/P

PV = nRT

Where P = pressure in atm, Pa, torr, mmHg or Nm-2; V = volume in cm3, dm3, ml or L; T = temperature in Kelvin; n = number of moles of gas in mol; R = molar gas constant = 0.082 dm3atmK-1mol-1

7 0
4 years ago
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
lbvjy [14]
-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s 

                                                                           = <span> 3,600 meters .</span>
___________________________________

The formula that has all of this in it is the formula for 
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is 
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds. 

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !  

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with. 
5 0
3 years ago
An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container ha
hichkok12 [17]

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

F_D =F_g

By definition we know that the Drag force is defined as

F_D= \frac{1}{2} C_d \rho A V^2

Where,

C_d = Drag coefficient

\rho =Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

F_g = (m_1 +m_2) g

Where,

m_1 =Mass of sphere

m_2 =Mass of unknown object

Equating the two equations we have to

(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2

Re-arrange for m_2,

m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1

Our values are given by,

C_d = 0.5

\rho = 1.22Kg/m^3

V = 66.7m/s

m_1 = 3Kg

d= 32.7*10^{-2}m

r = 16.35*10^{-2}m

Replacing in the equation we have,

m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3

m_2 = 8.62Kg

<em>Therefore the mass of unknown object is 8.62Kg</em>

4 0
3 years ago
If you are inside a car and the car is struck by lightning, it is best to stay inside because the electric field inside the car
Contact [7]

Answer:

Get out if the car

Explanation:

Get out of the car because the car could explode by over eclectic power

5 0
3 years ago
Read 2 more answers
Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours
Nostrana [21]

Answer:

Approximately 116\; \text{miles} for the train from Boston to NYC Penn Station.

Approximately 105\; \text{miles} for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}.

\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}.

Calculate average speed of each train:

\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}.

\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}

Assume that it takes a time period of t for the trains to pass by each other after departure. Distance each train travelled would be:

s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t.

s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t.

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}.

Rearrange and solve for t:

(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}.

\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}.

Distance each train travelled in t = (539 / 279)\; \text{hour}:

\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}.

\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}.

8 0
2 years ago
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