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elena-14-01-66 [18.8K]
4 years ago
10

Consider a completely elastic head-on collision between two particles that have the same mass and the same speed. What are the v

elocities of the particles after the collision?
Physics
1 answer:
motikmotik4 years ago
8 0

Answer:

The magnitude of their velocities will be the same but their direction will be reversed.

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A 1 in diameter solid round bar has a groove 0.1 in deep with a 0.1 in radius machined into it. The bar is made of AISI 1040 CD
Maru [420]

The answer & explanation for this question is given in the attachment below.

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4 years ago
a 63 kg object needs to be lifted 7 meters in a matter of 5 seconds. approximately how much horsepower is required to achieve th
Natali [406]
Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:

W = Fd

where:

F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration

Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.

W = Fd = mad = 63(9.81)(7) = 4326.21 Joules

Now that we have work, we can calculate power.

P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts

Convert watts to horsepower (1 hp = 745.7 watts)

P = 865.242 watts (1hp/745.7 watts) = 1.16 hp

3 0
3 years ago
Read 2 more answers
Determine if the object would move or not
lorasvet [3.4K]
I don't see a picture
4 0
3 years ago
When a particle of charge q moves with a velocity v⃗ in a magnetic field B⃗ , the particle is acted upon by a force F⃗ exerted b
melomori [17]

Answer:

3.8\cdot 10^{-16}N

Explanation:

For a charged particle moving perpendicularly to a magnetic field, the magnitude of the force exerted on the particle is:

F=qvB

where

q is the magnitude of the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

In this problem, we have

q=e=1.6\cdot 10^{-19} is the charge of the electron

v=6.0\cdot 10^6 m/s is the velocity

B=4.0\cdot 10^{-4}T is the magnetic field

Substituting into the formula, we find the force:

F=(1.6\cdot 10^{-19} C)(6.0\cdot 10^6 m/s)(4.0\cdot 10^{-4}T)=3.8\cdot 10^{-16}N

8 0
4 years ago
In 0.414 s, a 8.86-kg block is pulled through a distance of 4.36 m on a frictionless horizontal surface, starting from rest. The
Lena [83]

Answer: x = 0.957 m

Explanation:

F = ma

d = ½at²

a = 2d/t²

F = m2d/t²

F = kx

kx = m2d/t²

x = 2md/kt²

x = 2(8.86)(4.36) / (471(0.414²))

x = 0.957036...

3 0
3 years ago
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