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2 years ago

Can you help me? Thanks! 10+ points!

Physics

2 answers:

kykrilka [37]2 years ago

7 0

If u need any other help I’d help ! Honestly the person above me answered but feel free to message me if u need help ! I don’t really need the points !

Elis [28]2 years ago

4 0

Explanation:

velocity: average velocity equals Displacement /change in time.

Kinetic energy: K=1/2mv^2 (m = Mass) ( v = Velocity).

speed using the kinetic energy is also KE =1/2mv^2

(M in Kg) (V in m/s)

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The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21

andrezito [222]

Answer:

Explanation:

Given

average speed of train $(v_{avg})=216 kmph\approx 60 m/s$

Maximum acceleration=0.05g

Now centripetal acceleration is

$a_c=\frac{v^2}{r}$

$0.05\times 9.8=\frac{60^2}{r}$

r=7346.93 m

(b)Radius of curvature=900 m

therefore $a_c=\frac{v^2}{r}$

$v=\sqrt{a_cr}$

$v=\sqrt{0.05\times 9.8\times 900}$

$v=\sqrt{441}=21 m/s$

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3 years ago

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0

3 years ago

A visual display of data or information is a

Artist 52 [7]

A visual display of data or information is called a graph. There are many types of graphs. These can include pie graphs, bar graphs, and many more. Graphs are useful, because they show you visually data which is helpful to many. Hope this helped

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3 years ago

A train slows from 60m/s to 20m/s in 50s

Dovator [93]

Answer:

a = -4/5 m/s^2

Explanation:

Acceleration = change in velocity / time

change in velocity = final velocity - initial velocity

a = (20 m/s - 60 m/s) / 50 s

a = -40 m/s / 50 s

a = -4/5 m/s^2

hope this helps! <3

7 0

1 year ago

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