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TEA [102]
3 years ago
7

How do I solve for time if I have initial velocity and final speed?

Physics
1 answer:
xeze [42]3 years ago
4 0
Final speed = initial speed + (acceleration x time)

(final speed - initial speed) = acceleration x time

Time = (final speed - initial speed) / acceleration
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A golf ball is hit horizontally off the edge of a 30 m high cliff and lands a distance of 25 m from the edge of the cliff. What
Ratling [72]

Answer:

V₀y = 0 m/s

t = 2.47 s

V₀ₓ = 61.86 m/s

Vₓ = 61.86 m/s

Explanation:

Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL

<u>V₀y = 0 m/s</u>

For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = 30 m

g = 9.8 m/s²

t = time to hit the ground = ?

Therefore,

30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

t² = 30 m/4.9 m/s²

t = √6.122 s²

<u>t = 2.47 s</u>

For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,

V₀ₓ = Xt

where,

V₀ₓ = Initial vertical Velocity = ?

X = Horizontal Distance = 25 m

Therefore,

V₀ₓ = (25 m)(2.47 s)

<u>V₀ₓ = 61.86 m/s</u>

<u></u>

Due, to uniform motion in horizontal direction:

Final Vertical Velocity = Vₓ = V₀ₓ

Vₓ = 61.86 m/s

4 0
3 years ago
A computational model predicts the maximum potential energy a roller coaster car can have given its mass and its speed at the lo
andreyandreev [35.5K]

Answer:

The prediction for its maximum potential energy is 109,375 J

Explanation:

Given;

mass of the coaster car, m = 350 kg

speed of the coaster car at the lowest point, v = 25 m/s

The coaster car will have maximum kinetic energy at the lowest point and based on law of conservation of mechanical energy, the maximum kinetic energy of the coaster car at the lowest point will be equal to maximum potential energy at the highest point.

K.E_{max} = P.E_{max}

K.E_{max} = \frac{1}{2} mv^2\\\\K.E_{max} = \frac{1}{2} (350)(25)^2\\\\K.E_{max} =109,375 \ J

Thus, P.E_{max} = 109,375 \ J

Therefore, the prediction for its maximum potential energy is 109,375 J

3 0
3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

d = v*t

Where:

v: is the tangential speed of the disk

t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s    

Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0
2 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7
Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I=\frac{P_{avg}}{A}

I=\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}

I=7.296\times 10^{-6} W/m^2

Amplitude of electric field at receiver end

E_{max}=\sqrt{2I\mu _0c}

Amplitude of induced emf

=E_{max}d

=\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75

=17.591\times 10^{-2}=0.1759 v

7 0
3 years ago
Basalt, formed from magma, contains iron-rich minerals. Why is this important?
Svetach [21]
Iron is a magnetic metal, and it is essential to the Earth's magnetic field!
Essentially it "records" (stores the information, maintains) the direction and orientation of the magnetic field.

Among others, the magnetic field protects the Earth from dangerous cosmic rays. 
5 0
3 years ago
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