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Step2247 [10]
3 years ago
10

AYe PeEpS iF yOu GiVe Me A wEriD aNsWeR ImA rEeEeeEeEpOrT iT :D

Physics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

Tt tT

tT tt

Explanation:

the rest I pretty sure don't understand

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A marble, rolling with speed of 20m/sec rolls off the edge of the table that is 180m high (g=10m/sec2), find time taken to drop
natali 33 [55]

Answer:

<em>Choice: c. 6sec</em>

Explanation:

<u>Horizontal Launch </u>

When an object is thrown horizontally with a speed (v) from a height (h), it describes a curved path ruled by gravity until it finally hits the ground.

The horizontal component of the velocity is always constant because no acceleration exists in that direction, thus:

v_x=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

Where g=10 m/s^2

To calculate the time the object takes to hit the ground, we use the same formula as for free-fall, since the time does not depend on the initial speed:

\displaystyle t=\sqrt{\frac{2h}{g}}

The marble rolls the edge of the table at a height of h=180 m, thus:

\displaystyle t=\sqrt{\frac{2*180}{10}}

\displaystyle t=\sqrt{36}

t = 6 sec

Choice: c. 6sec

6 0
3 years ago
Read 2 more answers
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
3 years ago
n an object. one force is 3N to the east and the other force is 9n to the west. what is the net force acting on the object ​
cricket20 [7]

Answer:

-6N

Explanation:

The force to the east is acting in the positive x-direction therefore it is positive. The force to the east is in the negative x-direction therefore it is negative. The net force is just the sum of the two so 3-9=-6

4 0
2 years ago
If you have any questions, let me know​
klasskru [66]

Answer:

The red car would experience the greatest acceleration.

Explanation:

Newton says that Force equals mass times acceleration or F = ma

We get a = F/m

If we want the greatest acceleration or a, mass or m must be the lowest.

6 0
3 years ago
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