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Finger [1]
3 years ago
11

A particle has velocity v⃗1 v → 1 as it accelerates from 1 to 2. What is its velocity vector v⃗2 v → 2 as it moves away from poi

nt 2 on its way to point 3?

Physics
1 answer:
Jobisdone [24]3 years ago
8 0

Answer:

The velocity of the particle will be downward.

Explanation:

Given that,

The velocity of a particle is v₁. It is accelerated from 1 to 2.

If it moves away from point 2 on its way then

We need to find the velocity of particle

According to figure,

A particle moves downward from 1 to 2 with the velocity v₁ and after that the particle moves downward from 2 to 3 with the velocity v₂.

Hence, The velocity of the particle will be downward.

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A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct
-Dominant- [34]

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e  \dfrac{m}{2} .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v

The velocity of the other fragment immediately following the explosion is v .

4 0
3 years ago
Isostatic rebound
svp [43]
I think it is B the rise of the land
4 0
3 years ago
In 1967, new zealander burt munro set the world record for an indian motorcycle, on the bonneville salt flats in utah, with a ma
Fed [463]
Draw a velocity-time diagram as shown below.

Because a velocity of 26.82 m/s is attained in 4.00 s from rest, the average acceleration is
a = 26.82/4 = 6.705 m/s²
The time required to reach maximum velocity of 82.1 m/s is
t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s

The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
    = (1/2)*(6.705 m/s²)*(12.2446 s)²
    = 502.64 m

Answer:
The time required to reach maximum speed is 12.245 s
The distance traveled during the acceleration phase is 502.6 m

3 0
3 years ago
Which statements accurately describe Ernest Rutherford’s experiment? Check all that apply.
Dimas [21]

Answer:

Option (1), option (4) and option (5)

Explanation:

The main observations of Ernest Rutherford's experiment are given below:

1. most of the positively charged particles pass straight, it means there is an empty space in the atom.

2. Very few positively charged particles retraces their path.

So,

The positively charged particles were deflected because like charges repel, that means they are deflected by protons.

Almost all the positively charge concentrate in a very small part which is called nucleus.

7 0
4 years ago
Michael is biking on a trail and is accelerating at a rate of 1.2 m/s/s for 15 seconds. He began this part of his ride with a ve
Serhud [2]

Answer:

Michael's final velocity is 19.62 m/s.    

Explanation:            

We can find the final velocity of Michael by using the following kinematic equation:

v_{f} = v_{0} + at   (1)    

Where:

v_{f}: is the final velocity =?

v_{0}: is the initial velocity = 1.62 m/s

a: is the acceleration = 1.2 m/s²

t: is the time = 15 s

By entering the above values into equation (1) we have:

v_{f} = 1.62 m/s + 1.2 m/s^{2}*15 s

v_{f} = 19.62 m/s

Therefore, Michael's final velocity is 19.62 m/s.

I hope it helps you!                                            

7 0
3 years ago
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