Answer:
The correct answer is 4R1
Explanation:
According to the given scenario ,the radius of the tightest curve on the same road without skidding is as follows:
As we know that
Centeripetal Acceleration is
= v^2 ÷ r
In the case when velocity becomes 2 times so the r would be 4 times
So, the radius of the tightest curve on the same road without skidding is 4R1
Answer:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards
</em>
Explanation:
<u>Vertical Throw
</u>
When an object is thrown upwards, it describes a special type of motion ruled only by gravity.
When the ball is launched, it has its maximum speed upwards. The acceleration of gravity is always the same because it's a constant value near our planet's surface. The object starts to lose speed since the acceleration of gravity is pointed downwards and makes the object stop in the mid-air at its maximum height, where the speed is zero. Then, the object starts to fall and regain speed, this time downwards until it reaches back the launching point at the very same speed it was launched, but in the opposite direction.
The time it takes to reach its maximum height is the same it takes to return to the catching point, 2 seconds later.
With all these concepts in mind, we state that:
<em>D. The acceleration after it leaves the hand is 10 m/s/s downwards </em>
The other options are not correct because:
A. The acceleration is never upwards
B. The acceleration is never 0
C. Both times are equal
Answer:
Incomplete question. Complete question is: An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. Determine the angle through which the drill rotates by this point.
The answer is : Δ θ = 1 rad
Explanation:
Ok, so the condition involves the centripetal acceleration and the tangential acceleration, so let’s start by writing expressions for each:
Ac= centripetal acceleration At= tangential acceleration
Ac = V² / r At = r α
Because we have to determine the angle ultimately, therefore we should convert the linear velocity into angular velocity in the expression for centripetal acceleration
V = r ω
Ac = (r ω)² / r = r² ω² / r
Ac = r ω²
now that we have expressions for the centripetal and tangential acceleration, we can write an equation that expresses the condition given: The magnitude of the centripetal acceleration is twice the magnitude of the tangential acceleration.
Ac = 2 At
That is,
r ω² = 2 r α
it is equivalent to;
ω² = 2 α
now we have the relation between angular speed and angular acceleration, but we also need to determine the angular displacement as well. Therefore choose a kinematics equation that doesn’t involve time because time is not mentioned in the question. Thus,
ω² – ω°² = 2 α Δ θ
such that ω° = 0
and ω² = 2 α
therefore;
2 α - 0 = 2 α Δ θ
2 α = 2 α Δ θ
So the angle will be : Δ θ = 1 rad
Acceleration equals 24 km/s
Average equals 396km/s
