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hichkok12 [17]
2 years ago
9

- A certain mass is suspended at one end of a spring of spring-constant of 32,000gm/s2. The

Physics
1 answer:
babunello [35]2 years ago
6 0

Answer: I am pretty sure it is (b) what is the value of mass suspended at the end of the spring.

Explanation:

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Please help!! I will give brainliest!
grandymaker [24]

Answer:

Depends.

Explanation:

Whether the object is going left or right, the speed will stay the same until friction eventually stops it. <em>However, </em>if, for example, we're talking about an object going straight before veering right, then yes, speed <em>does</em> matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.

So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0
3 years ago
If the wavelength is 10 m and the frequency is 5 Hz, what is the wave speed?
timama [110]
V = f x wavelength
V = 5 x 10
V = 50m/s
4 0
2 years ago
If you put a drop of food coloring in water and watch the drop disperse, is entropy increasing or decreasing.
den301095 [7]

Answer:

Entropy is increasing. Entropy is decreasing.

Explanation:

The Entropy doesn't change.

4 0
3 years ago
A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc
Hoochie [10]

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v=  0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

a=- \dfrac{20^2}{2\times 30} \ m/s^2

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be  - 6.667 m/s² .

6 0
3 years ago
Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a
mezya [45]

Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

q= +2e

The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

V= 9×10^9×(5.67×10^-14)= 5101.35v

3 0
3 years ago
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