In the horizontal direction, the forces acting on the person are
• friction with magnitude <em>f</em>, opposing motion, and
• the horizontal component of the pulling force (itself with mag. <em>p</em> ) with mag. <em>p</em> cos(30º), in the direction of motion.
There is no friction in the vertical direction, so we omit any discussion of the vertical forces.
By Newton's second law, we then have
<em>p</em> cos(30º) - <em>f</em> = <em>m</em> <em>a</em> cos(30º)
where <em>m</em> is the person's mass, and <em>a</em> is their acceleration so that <em>a</em> cos(30º) is the magnitude of the horizontal component of acceleration. The person is pulled by a force of <em>p</em> = 401 N, so solve for <em>f</em> :
(401 N) cos(30º) - <em>f</em> = (53 kg) (0.59 m/s²) cos(30º)
<em>f</em> ≈ 320 N
Answer:
V = -0.4 m/s
Explanation:
Given that,
Mass of cart 1, m₁ = 3 kg
Speed of cart 1, v₁ = 2 m/s
Mass of cart 2, m₂ = 2 kg
Speed of cart 2, v₂ = -4 m/s (left)
aWe need to find the velocity of the carts after the collision if they stick together. Let the common speed be V. Using the law of conservation of momentum to find it such that,
Put all the values,
So, the velocity of the carts after the collision is equal to 0.4 m/s to the left.
C - velocity
When you look at a point it is saying it goes this many meters in this amount of time. This is speed which means the slope is velocity
Answer: equal to 3 m/s
Explanation:
Speed of golf ball will be equal to 3 m/s because in Perfect Elastic Collision Energy is conserved .
So speed of golf ball will be same in order to Satisfy
Initial Kinetic Energy =Final Kinetic Energy
Considering Bowling ball remains at rest after collision other wise some energy will also be acquired by bowling ball which automatically decreases the amount of Kinetic Energy of golf ball resulting its speed to decrease by some extent.
It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.
To find the answer, we need to know about the third law of Kepler.
<h3>What's the Kepler's third law?</h3>
- It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
- Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
- The time period of geosynchronous orbit is 24 hours or 1440 minutes.
- As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
- If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
- a1= (T1/T2)⅔×a2
= (1440/90)⅔×6780
= 43,090 km
- Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km
Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.
Learn more about the Kepler's third law here:
brainly.com/question/16705471
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