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ivanzaharov [21]
3 years ago
10

A member of the Penn State Women's Basketball team performs a jumping movement simulating that used in a jump shot. The time fro

m take-off to landing (total time) is measured to be 0.8 seconds. Compute the following,
a) The vertical displacement of her center of mass (height of jump) Vertical Displacement of Center of Mass = __________

b) Her initial vertical velocity at the instant of takeoff. Initial Velocity at takeoff = __________
Physics
1 answer:
Shkiper50 [21]3 years ago
5 0

A) Starting from rest, we have the entire cycle determined by 0.8s.

If we assume a constant movement, half of that time is when it reaches the highest point, that is, in 0.4s.

The distance as a function of speed and acceleration is given by,

d=v_it+\frac{1}{2}at^2

At the initial point the speed is zero and the acceleration is equivalent to gravity.

d= 0+ \frac{1}{2}9.8*0.4^2

d=0.784m

B) When returning to the ground, the final speed is zero. Therefore, the equation that relates velocity to acceleration is given by,

V_f = V_i+at

0 = V_i -9.8(0.4)

V_i = 3.92m/s

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Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How
grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (\Delta s), in meters, can be determined by the following expression:

\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2} (1)

Where:

v_{o} - Initial velocity, in meters per second.

t - Time, in seconds.

a - Acceleration, in meters per square second.

If we know that v_{o} = 5\,\frac{m}{s}, t = 6\,s and a = 2\,\frac{m}{s^{2}}, then the box displacement after 6 seconds is:

\Delta s = 66\,m

The box displacement after 6 seconds is 66 meters.

5 0
3 years ago
We know that the Moon revolves around Earth during a period of 27.3 days. The average distance from the center of Earth to the c
PtichkaEL [24]

Answer:

Explanation:

This is a circular motion questions

Where the oscillation is 27.3days

Given radius (r)=3.84×10^8m

Circular motion formulas

V=wr

a=v^2/r

w=θ/t

Now, the moon makes one complete oscillation for 27.3days

Then, one complete oscillation is 2πrad

Therefore, θ=2πrad

Then 27.3 days to secs

1day=24hrs

1hrs=3600sec

Therefore, 1day=24×3600secs

Now, 27.3days= 27.3×24×3600=2358720secs

t=2358720secs

Now,

w=θ/t

w=2π/2358720 rad/secs

Now,

V=wr

V=2π/2358720 ×3.84×10^8

V=1022.9m/s

Then,

a=v^2/r

a=1022.9^2/×3.84×10^8

a=0.0027m/s^2

3 0
3 years ago
20 points and brainliest‼️‼️‼️‼️
Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

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