3 years ago

Showing how a projectile motion is parabolic

1 answer:

4 0

Answer:

In the equation, d=vi*t+1/2*a*t^2, you can see that this is a quadratic with respect to time, so it follows a parabolic motion

Explanation:

You might be interested in

What is the kinetic energy of the ball just before it hits the ground?

Alex_Xolod [135]

Answer:

$0.5mv^{2}$

Explanation:

Kinetic energy, $KE= 0.5mv^{2}$

while potential energy, PE= mgh where m is the mass, v is the velocity of the ball, g is acceleration due to gravity and h is the height.

The question requires kinetic energy hence we use the $KE= 0.5mv^{2}$

Since the figures are not given, we maintain that $KE= 0.5mv^{2}$

4 0

3 years ago

Derek runs 4 laps around the track. If each lap around the track is 0.25 miles long, and he starts and stops in the same locatio

Sergeeva-Olga [200]

1.00 1 mile 1 mile 1.00

5 0

4 years ago

Read 2 more answers

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$