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katovenus [111]
3 years ago
7

Suppose you know a golf ball's horizontal velocity and the time it has traveled through the air. how could you calculate how far

the ball traveled
Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Given the time and the horizontal velocity, we can simply compute for the distance how far the ball travelled using the formula:

distance = velocity * time

 

<span>Since velocity is in units of m/s and time is seconds, therefore we can directly get a unit in meters.</span>

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The purpose of the little Albert experiment?
HACTEHA [7]

Answer:

The aim of Watson and Rayner was to condition a phobia in an emotionally stable child.

Explanation:

Does this help?

3 0
3 years ago
A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Lilit [14]

Answer:

v = 1.08 m/s

Explanation:

What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

The decrease in PE is

d = 80.0cm * 1 / 1000m = 0.80m

h = 0.80 m /2 = 0.40 m

ΔPE = m*g*h

ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m

ΔPE = 0.1176 J

The moment of inertia of the assembly is

I = 1/12*m*L² + (m1 + m2)*(L/2)²

I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²

I = 0.032 kg·m²

KE = ½Iω²

0.1176 J = ½ * 0.032kg·m² * ω²

ω = 2.71 rad/s

v = ωr = 2.71 rad/s * 0.400m

The linear velocity

v = 1.08 m/s

3 0
3 years ago
A car traveling at a constant speed travels 175 miles in 4 hours. How many feet will the car travel in 10 minutes?
8_murik_8 [283]
The answer would be 70. I got my answer from www.iun.edu
8 0
3 years ago
If we increase the distance traveled when doing work , and keep all other factors the same, what will happen? (2 points) The amo
Solnce55 [7]
I think the amount of force will decrease and the amount of work will increase
4 0
4 years ago
Read 2 more answers
A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st
sp2606 [1]

Answer:

A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2  ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck :       m1 = 2000 Kg           v1 = 5 m/s                u1 = 0

car    :        m2 = 1000 kg           v2 = 0                      u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = \frac{2000}{1000} x 5

    = 10 m/s

Hence after collision, car moves at a velocity of 10 m/s

3 0
3 years ago
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