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BartSMP [9]
3 years ago
5

3. A 6 kg block moving to the right at 4 m/s collides with and sticks to a stationary block of unknown mass. If the two blocks m

ove off to the right at 3 m/s, what was the mass of the stationary block?
Physics
1 answer:
Sphinxa [80]3 years ago
6 0

Answer: 2kg

Explanation:

This problem is a textbook conservation of momentum problem. The intial momentum is equal to the final momentum. For the initial state of each block, only the first one was moving. Then they both combine to move together.

Pi = Pf

with p = mv

(6kg)*(4m/s) = (6kg+xkg)(3m/s)

Let x equal the unknown mass of block 2

24 = 18 + 3x

6 = 3x

x = 2kg

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Amy has a mass of 30 kg, and she is riding a skateboard traveling 5 m/s. What is her momentum?
Flauer [41]
Hey there!

So we know that m*v=P.
And in this question m=30
v=5 m/s
P = 30*5 Kgm/s
P = 150 Kgm/s

So, your final answer is 150 Kg.m/s

Hope this helps! :)
5 0
3 years ago
Read 2 more answers
Phosphorus (P) is an element with an atomic number of 15 and an atomic mass of 31. How many neutrons are in an atom of phosphoru
BaLLatris [955]

(31-15 = 16).

Explanation:

the element phosphorus (P) has an atomic number of 15 and a mass number of 31. Therefore, an atom of phosphorus has 15 protons, 15 electrons, and 16 neutrons

5 0
3 years ago
The atomic number of an element is found by?
azamat

Calculate the number of neutrons. Now you know that atomic number = number of protons, and mass number = number of protons + number of neutrons. To find the number of neutrons in an element, subtract the atomic number from the mass number.

6 0
2 years ago
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An 80.0 N force accelerates a 6.0 kg object from 4.0 m/s to 8.0 m/s. What is the
kolezko [41]

Answer:

24kgm/s

Explanation:

Given parameters:

Force  = 80N

Mass of object  = 6kg

Initial velocity  = 4m/s

Final velocity = 8m/s

Solution:

Impulse  = ?

Solution:

The impulse on a body is its change in momentum.

    Impulse  = m (v  - u )

m is the mass

v is the final velocity

u is the initial velocity

  Now insert the parameters and solve;

          Impulse  = 6 (8 - 4) = 24kgm/s

8 0
2 years ago
A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 35 Hz. Determine the maximum power that the shaft can trans
Phantasy [73]

Answer:

P=58.3kW

Explanation:

Given data

Length L=2.5 m

Radius R=d/2=30/2 = 15 mm

Torque based on allowable stress

Allowable shear stress τ=50 Mpa

Allowable torque  T=(π/2)τc³

T=\frac{\pi }{2}(50*10^{6} )(0.015)^{3} \\ T=264.94N.m

Torque based on allowable angle of twist

Allowable Angle of twist

Ф=7.5°

Ф=7.5×(π/180)=130.90×10⁻³ rad

Allowable torque

T=(GJФ)/L

T=(G(π/2)c⁴)Ф)/L

T=(πGc⁴Ф)/2

T=\frac{\pi (77.2*10^{9} )(0.015)^{4}(130.90*10^{-3} ) }{2}\\ T=265.07N.m

Maximum Power Transmitted

Maximum power transmitted is given by

P=2\pi fT\\P=2\pi (35)(265.07)\\P=58262.386watt\\P=58.3kW

6 0
3 years ago
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