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EleoNora [17]
3 years ago
5

A motorcycle traveling at 25 m/s accelerates ya a rate of 7.0 m/s2 for 6.0 seconds. What is the final velocity of the motorcycle

?
Physics
1 answer:
frutty [35]3 years ago
4 0

First write down all your known variables:

vi = 25m/s

a = 7.0m/s^2

t = 6.0s

vf = ?

Then choose the kinematic equation that relates all the variables and solve for the unknown variable:

vf = vi + at

vf = (25) + (7.0)(6.0)

vf = 67m/s

The final velocity of the motorcycle is 67m/s.

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A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance
FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0
3 years ago
How to find i1, i2,i3
MrRissso [65]

to find i1, i2, and i3 we need to find the total current.

to find the total current, you need to find the total resistance

you're already given the total voltage, Vs

to find Rtotal, start from the resistors furthest from the voltage source.

R3 and R4 are in series so

Rtotal= R3+R4 = 6+3 = 9 ohms

9 ohms is now in parallel with R2 so,

Rtotal= (\frac{1}{R3+R4}) ^{-1}\\ + (\frac{1}{R2}) ^{-1})^-1= (1/18)^-1 +( 1/9)^-1 = 6 ohms

6 ohms is in series with R1 so

Rtotal=  4+6=10 ohms

itotal= (\frac{Vtotal}{Rtotal})

= 120 v/10 ohms = 12 A

i total = i1 because all the current flows through it

i1= 12A

so the current splits into i2 and i3 and the amount of current that flows through a branch depends on the total resistance in each branch.

we already calculated the resistance in the R3+R4 & R2 branch as 6 ohms

since r3 and r4 are in series, the same current will flow through them

r3+r4 = 9 ohms

r2= 18 ohms

so the current in r2 will be half that of R3 & R4 (V=IR)

using the current divider rule

Ix = Itotal * \frac{Rtotal}{Rx}

i2= 12A x (6 ohms/18 ohms)= 4 A

i3= 12A x (6 ohms/9 ohms) = 8 ohms

6 0
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
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Alexxandr [17]
You need to post a picture so someone can help
6 0
3 years ago
Identify the energy transformations that take place in an electrical fan heater
Sergio [31]
The energy goes from electric energy and gets converted into thermal energy.
3 0
3 years ago
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