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Mkey [24]
3 years ago
5

Can someone pls help me?!!!! i need it asap!! i’m very stressed

Physics
1 answer:
Mariulka [41]3 years ago
6 0

Answer:

Acceleration

Explanation:

Its speed or velocity change

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To find the number of neutrons in an atom you would subtract?
Ghella [55]
You need to subtract the atomic number from the atomic mass to find the # of neutrons. 
8 0
3 years ago
A particle whose speed is 50 m/s moves along the line from A(2,1) to B(9,25). Find its velocity vector in the form of ai +bj.
Kitty [74]

To write it in the form ai + bj, we need to find a and b which are:

a = x component

b = y component

 

length of x = 9 – 2 = 7

length of y = 25 – 1 = 24

 

tan θ = 24 / 7

θ = 73.74°

 

a = (50 m/s) cos 73.74

a = 14 m/s

 

b = (50 m/s) sin 73.74

b = 48 m/s

 

Hence,

14i + 48j

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3 years ago
Definition of mass<br> definition of density<br> definition of physical properties
butalik [34]

Answer:

Mass of a body is define as the total quantity of matter contained in it .

Density is the mass per unit volume in any object .

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6 0
3 years ago
What are the three types of motion
Cloud [144]
Rectilinear Motion, Circular Motion and Periodic Motion.
7 0
3 years ago
4. A massless spring hangs from the ceiling, and a mass is hung from the bottom of it. The mass is supported
MrRissso [65]

Answer:

The correct option is C: 0.31 s.

Explanation:

When the mass is then suddenly released we have:

F = k\Delta y

Where:

F is the force

k: is the spring constant

Δy: is the spring displacement

Since the tension in the spring is zero, the force is the weight:

F = mg

Where:

m is the mass of the object

g is the gravity

mg = k\Delta y    (1)

The oscillation period of the spring is given by:

T = 2\pi \sqrt{\frac{m}{k}}    (2)

By solving equation (1) for "k" and entering into equation (2) we have:

T = 2\pi \sqrt{\frac{m}{\frac{mg}{\Delta y}}}

T = 2\pi \sqrt{\frac{\Delta y}{g}}

Since the spring will osclliates in a position between the initial position (when it is at rest) and the final position (when the mass is released and reaches the bottom), we have Δy = 2.5 cm = 0.025 m:

T = 2\pi \sqrt{\frac{0.025 m}{10 m/s^{2}}} = 0.31 s                   

Hence, the oscillation period is 0.31 s.

The correct option is C: 0.31s.

I hope it helps you!                                                                                                      

7 0
4 years ago
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