4. With a diameter that's 11 times larger than Earth's, ______

forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s

zhenek [66]

Answer:

345453-5676

its the right answer

5 0

3 years ago

Calculate the velocity of the object in the graph during the following time interval 0-6 seconds. Also, during which time interv

cestrela7 [59]

Its accelarating from 1-10 i know that because the line is going up. i hope i helped

3 0

3 years ago

a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s

Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

$h = v_0_yt + \frac{1}{2} gt^2$

where;

t is the time of motion
$v_0_y$ is the initial vertical velocity of the stone = 0

$h = \frac{1}{2} gt^2$

The time taken by the first stone to hit the ground is calculated as;

$t_1 = \sqrt{\frac{2h}{g} }$

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

$t_2 = \sqrt{\frac{2h}{g} } + 1$

$t_2 = t_1 + 1$

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"Your question is not complete, it seems be missing the following information;"

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0

3 years ago

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can

Sonja [21]

Answer:

Explanation:

a ) speed of passenger = circumference / time

= 2π R / Time

= 2 x 3.14 x 50 / 60

= 5.23 m /s

b )

centrifugal force = m v² /R

= (882 /9.8 ) x 5.23² / 50

= 77.47 N

Apparent weight at the highest point

real weight - centrifugal force

= 882 - 77.47

= 804.53 N

Apparent weight at the lowest point

real weight + centrifugal force

= 882 +77.47

= 959.47 N

c ) if the passenger’s apparent weight at the highest point were zero

centrifugal force = weight

mv² /R = mg

v² = gR

= 9.8 X 50

v = 22.13 m /s

d )

apparent weight

mg - mv² / R

= 882 - (882 / 9.8 )x 22.13²/50

= 882 + 882

= 1764 N

=

6 0

3 years ago

A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the

Alecsey [184]

Answer:

Part a)

$KE = 77.95 J$

Part b)

$L = 3.16 m$

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

$KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2$

so we will have

$KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2$

$KE = \frac{7}{10} mv^2$

$KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)$

$KE = 77.95 J$

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

$mgLsin\theta = KE$

$\frac{42}{9.81}(9.81)L sin36 = 77.95$

$L = 3.16 m$

Part c)

by equation of energy conservation we know that

$\frac{7}{10}mv^2 = mgL sin\theta$

so here we can see that distance L is independent of the mass of the sphere

7 0

3 years ago

4. With a diameter that's 11 times larger than Earth's, _______ is the largest planet.