How fast, in rpm, would a 5.8 kg, 22-cm-diameter bowling ball have to spin to have an angular momentum of 0.21 kgm2/s

Gwar [14]

Answer:

If iron, nickle, or cobalt is an answer choice, those are the three metals that are purely magnetic.

3 0

3 years ago

An acorn falls from a branch located 9.8 m above the ground. After 1 s of falling, the acorn's velocity will be 9.8 m/s downward

goldenfox [79]

Answer:

The acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

Explanation:

given information:

h =9.8

t =1 s

g = 9.8

the average speed

v = 1/2 gt²

= 1/2 (9.8) (1)²

= 4.8 m/s

the distance in 1s

h = v t

= 4.8 (1)

= 4.8 m

the acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

=

8 0

3 years ago

1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15

kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity= $\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h$

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 $m/s^2$

Distance covered by the car,s = 75 m

Initial velocity of the car , $v_i$ = 6 m/s

Final velocity of thre car , $v_f$ =?

Using third equation of motion:

$v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2$

$v_{f}=14.69 m/s$

The final velocity of the car at the end of 75 m is 14.69 m/s

8 0

3 years ago

A rabbit runs 28 m toward the left in 9 s; then the rabbit suddenly changes direction and runs 18 m toward the right in the next

Viefleur [7K]

Answer:

The answer is 3.111111.

Explanation:

It runs 28 m in the first 9 s, and 28 divided by 9 equals 3.1 and the one goes on forever.

8 0

3 years ago

What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg

scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

$F_c = F_g$

$\frac{mv^2}{r} = \frac{GmM}{r^2}$

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

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

At the same time we know that the period is equivalent in terms of the linear velocity to,

$T = \frac{2\pi}{\frac{v}{r}}$

$v = \frac{2\pi r}{T}$

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

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

$( \frac{2\pi r}{T})^2 = \frac{GM}{r}$

$T = \sqrt{\frac{4\pi^2 r^3}{GM}}$

Replacing we have,

$T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}$

$T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})$

$T= 1.74hour$

Therefore the correct answer is C.

7 0

3 years ago