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makvit [3.9K]
3 years ago
11

Will mark brainliest! Please help!

Physics
1 answer:
djverab [1.8K]3 years ago
3 0

Answer:

30 seconds

Explanation:

A = A02^-(t/hl)

--> ln(A/A0) = -(t/hl)ln2

solving for hl,

hl = -t x ln2 /ln(A/A0)

= -(60 min)xln2/ln(50/200)

= 0.5 min or 30 seconds

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what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t
kykrilka [37]
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

 
6 0
3 years ago
A bicyclist is travelling at 25 m/s when he begins to decelerate at -4m/s2 . How fast is travelling after 5 seconds
Umnica [9.8K]

Initial velocity (Vi) = 25 m/s

acceleration (a) = -4 m/s^{2}

time interval (t) = 5 sec

let us assume that final velocity after 5 sec be Vf

As acceleration is constant, we can apply the the equation of motion with constant acceleration i.e. V_{f} = V_{i} + at

Hence, V_{f} = 25 +(-4)(5) = 25 -20 = 5 m/s

so, the velocity of bicyclist will be 5 m/s after 5 sec

7 0
3 years ago
Which wave has a greater frequency
larisa86 [58]

Answer:

A I think

Explanation:

because what is the most frequency a because it has more frequency I think I'm not that sure

5 0
3 years ago
Read 2 more answers
A football punter accelerates a .55 kg football
Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the change in momentum of the football

\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.55 kg is the mass of the football

u = 0 is the initial  velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

5 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
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