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Korolek [52]
3 years ago
5

A rock of mass 170 kg needs to be lifted off the ground. One end of a metal bar is slipped under the rock, and a fulcrum is set

up under the bar at a point that is 0.65 m from the rock. A worker pushes down (perpendicular) on the other end of the bar, which is 1.9 m away from the fulcrum. What force is required to move the rock?

Physics
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

866.32 N

Explanation:

The diagram explains better.

Taking the total moment of forces, the sum of the clockwise moment of forces about the fulcrum must be equal to the sum of the anticlockwise moment of forces about the fulcrum:

F * (1.9 - 0.65) = 170 * 9.8 * 0.65

F * 1.25 = 1082.9

F = 1082.9/1.25

F = 866.32 N

A force of 866.32 N is needed to move the rock.

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Oooo that ones hard. ummm... idk i think we should just leave it to the experts ya know.
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3 years ago
A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r
cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

t = 160 years.

Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0
2 years ago
A home run is hit in such a way that the baseball just clears a wall 21m hgh located 130m from home plate the ball is hit at an
svlad2 [7]

Answer:

see the attachment

Explanation:

take coordinate system correctly. use formulas of projectile motion

Download pdf
3 0
2 years ago
HOW FAR DOES A UNICYCLE TRAVEL AT A SPEED OF 20 M/S FOR 15 SECONDS?​
astra-53 [7]

Given:-

  • Speed of the unicycle = 20 m/s
  • Time taken = 15 s

To Find: Distance travelled by the unicycle.

We know,

s = vt

where,

  • s = Distance travelled,
  • v = Speed &
  • t = Time taken.

Therefore,

s = (20 m/s)(15 s)

→ s = (20 m)(15)

→ s = 300 m (Ans.)

8 0
3 years ago
This speed is measured with respect to the space station the spacecraft was originally launched from. In interstellar space the
valentinak56 [21]

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2

Equation of motion

v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

5 0
3 years ago
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