ASAP ONLY CORRECT ANSWERS PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEE I RLLY NEED HELPPPPPPP

53. A recently discovered planet has a mass four times as great as Earth's and a radius twice as large as Earth's. What will be

marusya05 [52]

Answer:

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

Explanation:

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

here we have

$M = 4M_e$

$R = 2R_e$

we know that for earth we have

$g = 9.81 = \frac{GM_e}{R_e^2}$

now if the radius and mass is given as above

$g' = \frac{G(4M_e)}{(2R_e)^2}$

$g' = \frac{GM_e}{R_e^2}$

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

6 0

3 years ago

Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary

balandron [24]

A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
$W=mg$
where
m is the mass
$g= \frac{GM}{r^2}$ is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, $g=9.81 m/s^2$ , so Gary's weight is
$W=mg=9.81 m$ (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is
$R=6370 km$
So the distance of Gary from the Earth's center is
$r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m$

The Earth's mass is $M=5.97 \cdot 10^{24} kg$ , so the gravitational acceleration is
$g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2$

Therefore, Gary's weight at 300 km above Earth's surface is
$W' = mg' = 8.95 m$ (2)

If we compare (1) and (2), we find that Gary's weight has changed by
$\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91$
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.

6 0

3 years ago

Convection currents affect the movement of plates on the Earth’s crust. True or false.

GREYUIT [131]

True

I hope this helps and have a wonderful day filled with joy!!

8 0

3 years ago

A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0

Lubov Fominskaja [6]

Answer:

t = 7.5 s

Explanation:

The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,

s₁ = Vi t + (0.5)gt²

where,

s₁ = distance traveled by police car

Vi = Initial Velocity = 0 m/s

t = time taken

Therefore,

s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

s₁ = 4.9 t²

Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:

s₂ = Vt = V(t + 1)

where,

s₂ = distance traveled by car

V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s

Therefore,

s₂ = 27.78 t + 27.78

Now, we know that at the time of meeting:

s₁ = s₂

4.9 t² = 27.78 t + 27.78

4.9 t² - 270.78 t - 27.78 = 0

solving the equation and choosing the positive root:

t = 6.5 s

since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.

t = 6.5 s + 1 s

6 0

4 years ago

Which can be used as a lever ?

Svet_ta [14]

B long board because of the length

7 0

3 years ago