Can someone help me find the answer?

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

3 years ago

If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is

Harrizon [31]

Answer:

A.) 42.7 m/s

B.) 0.33 m/s^2

C.) 90 kg

Explanation:

A.) If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is his velocity?

Velocity = displacement/time

Velocity = 2560/60

Velocity = 42.67 m/s

B.) The Chevy S-10 started rounding at 10 meters per hour. What is the acceleration at 30 seconds on the highway?

Acceleration = velocity/time

Acceleration = 10/30

Acceleration = 0.33 m/s^2

C.) The S-10 has a force of 30 N. What is the mass of the car?

Force = mass × acceleration

30 = mass × 0.33

Mass = 30/ 0.33

Mass = 90 kg

8 0

2 years ago

Alex has to pick up a 20 N stack of documents and raise them 5 meters from the floor.

Serjik [45]

Answer:

1.Work done=100 joules

2.potential energy

3.kinetic energy

Explanation:

W=Force x distance

w= 20 N x 5 M

W = 100 JOULES

6 0

3 years ago

I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)

Helga [31]

Hello! I can help you with this!

4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.

5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.

If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.

4 0

2 years ago

Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the

tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s$

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m$

The displacement of the ball on the moon is 7.04 m

6 0

3 years ago