All categories

2 years ago

What is this question formula a=V²-U² /2S?

Physics

1 answer:

erik [133]2 years ago

7 0

Answer:

sorry I don't really know about that question.

Explanation:

You might be interested in

Why concave lens is called diverging lens

dangina [55]

When a parallel beam of light passes through a convex lens, the rays become farther from one another when the come out. This process of rays is called ''to diverge''. The concave lens makes rays of light diverge, so it is called diverging lens.

6 0

2 years ago

you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?

laiz [17]

At the same speed because it will slow down as it approaches the peak then speed up as it goes down again

it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something

8 0

2 years ago

The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?

Eduardwww [97]

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

6 0

2 years ago

The half-life of a certain radioactive substance is 12 hours. There are 19 grams present initially. a. Express the amount of su

neonofarm [45]

Answer:

(a) $N=19\times e^{-\lambda t}$

(b) 15 hours

Explanation:

half life, T = 12 hours

No = 19 g

(a) Let N be the amount remaining after time t.

Let λ be the decay constant.

$\lambda =\frac {0.6931}{T}$

The equation of radioactivity used here is given by

$N=N_{o}e^{-\lambda t}$

$N=19\times e^{-\lambda t}$

(b) N = 8 gram

Substitute the values in above equation

$\lambda =\frac {0.6931}{12}$

λ = 0.0577 per hour

So, $8=19\times e^{-0.577t}$

$e^{-0.0577t}=0.421$

Take natural log on both the sides

- 0.0577 t = - 0.865

t = 15 hours

4 0

2 years ago

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

jeka57 [31]

Let $F_{1}=F_{2}=F$ .

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$ .

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$ , but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$ , solving for $F$ : $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$

7 0

2 years ago

What is this question formula a=V²-U² /2S?​

What is this question formula a=V²-U² /2S?