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galben [10]
3 years ago
6

An object completes one and half revolution of a circle of radius R calculate the displacement and distance

Physics
1 answer:
Papessa [141]3 years ago
8 0

Answer:

Distance is path length covered by particle. When particle moves along half circle, it covers half the circumference therefore distance covered is (2×pi×r)/2 = pi× r. ... Hence displacement is equal to diameter or 2 times the radius of circle.

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Which of the following elements is in the same period as phosphorus?
Arte-miy333 [17]
The answer is B. magnesium I am pretty sure
5 0
3 years ago
Which of the following is a metal? Calcium (Ca) Iron (Fe) Sodium (Na) all of these
ANTONII [103]
Iron (Fe) or at least it can be.

8 0
3 years ago
Read 2 more answers
A toy doll and a toy robot are standing on a frictionless surface facing each other. The doll has a mass of 0.2 kg, and the robo
Natali5045456 [20]

Answer:

<h3>1.43m/s²</h3>

Explanation:

According to newtons second law.

F = mass * acceleration

If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg

Force applied = 1N

acceleration = Force/mass

Substitute the values and get acceleration

acceleration = 1/0.7

acceleration = 1.43m/s²

Hence the magnitude of the acceleration of the robot is 1.43m/s²

3 0
2 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
A cannon ball is fired directly upward with a velocity of 160 m/s. How long does it take to fall back to the ground? s How fast
Andrej [43]
To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
6 0
3 years ago
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