Which planets have rings

a car advertisement states that a certain car can accelerate from rest to 70km/h in 7 seconds. find the cars acceleration

ELEN [110]

<span>a = (v2 - v1)/t= acceleration formula

a = (70 - 0)/7
a = 10 km/hr/sec
-----
It's better to use as few units as possible.
10 km/hr = 10 km*1000 m/km/(1 hr*3600 sec/hr) = 25/9 m/sec
a= 25/9 m/sec/se</span>

4 0

4 years ago

A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh

lianna [129]

Answer: 3.21 N

$Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}$

$\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg$

For weight, we will multiply by $g=9.8 m/s^{-2}$

$weight= 0.328\times9.8=3.21\hspace{1mm}N$

Hence, the rock would weigh 3.21 N.

3 0

3 years ago

The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point

Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth, R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

w = 2π / T

w = 2π / 24*3600

w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

v1 = R*w

v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

π/2 ........... s

x ............ 1/5 s

x = π/2*5 = 18°

- The radius of the earth R' at point where θ = 18° from the equator is:

R' = R*cos(18)

R' = (6.37 * 10 ^6)*cos(18)

R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

v2 = R'*w

v2 = (6058230.0088)*(7.27 * 10^-5)

v2 = 440.433 m/s

5 0

3 years ago

which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given

Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>

Weight of Punch on Earth = 236 lb
Desired weight = 118 lb

The mass of Punch will be constant in every planet;

$W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}$

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is the mass of Earth = 5.972 x 10²⁴ kg
R is the Radius of Earth = 6,371 km

For Planet Tehar;

$g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g$

For planet Loput:

$g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g$

For planet Cremury:

$g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g$

For Planet Suven:

$g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g$

For Planet Pentune;

$g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g$

For Planet Rams;

$g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g$

The weight Punch on Each Planet at a constant mass is calculated as follows;

$W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb$

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0

2 years ago

A uniform rod of length 0.8 m and mass 1.4 kg, has two point masses at each end. The point mass on the left end has a mass 1.2 k

VladimirAG [237]

Answer:

Explanation:

1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m

5 0

3 years ago