Answer:
17.0 m s^-1
Explanation:
This question can be answered using suvat equations: (I'm assuming you know what these are)
s = 34.0 m
u = 0
v = ?
a = N/A
t = 4.00s
The only suvat equation that doesn't contain "a" is s = (u+v/2)t.
Make "v" the subject so v = 2s/t.
v = (2 x 34.0)/4.00 = 17 m/s
Answer:
option A.
Explanation:
given,
Both swings are at the same height
Start swinging at the same time
The length of the string of your friend's string is longer than yours.
Using the formula of the time period
The time period is directly proportional to Length.
If the length of the string is less time period will be less.
So, your time period is less than your friend's.
Hence, when you will complete your swing your friend will be slightly lower than you.
Hence, the correct answer is option A.
Answer:
1.8m/s^2
Explanation:
Since the two ropes are going up, their combined force is 105+115=220N. With a gravitational force of 186N, the force of the two ropes pulling up the will be 220-186=34N.
Now we need the mass of the bucket itself in order to find the acceleration of the bucket (remember that F=ma and m is needed to find a). Since gravitational acceleration is 9.8m/s^2 and F=186N, 186/9.8=18.97959184 kg for the mass of the bucket.
Now that we have the mass of the bucket, we can find the acceleration of the bucket. Since F=34N from earlier, 34N/18.97959184kg=1.791397849m/s^2=1.8m/s^2 is the acceleration of the bucket.
Therefore, 1.8m/s^2 is the correct answer.
Please mark brainliest!
Answer:
This was my best estimation of the answers
Answer:
Explanation:
Let 'F₁' and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.
Now, the horizontal and vertical components of these forces are:
As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,
Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.
At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.
Therefore, the net torque by the forces will be zero. This gives,
But,
Therefore,
We know,
∴