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KatRina [158]
2 years ago
13

90cm uniform lever has a load 30N suspended at 15cm from one of it's end. If the fulcrum is at the center of gravity. The force

that must be applied at it's other hand to keep it in horizontal equilibrium is.​
Physics
1 answer:
solniwko [45]2 years ago
3 0

Answer: F = 20 N

Explanation:

I will ASSUME that the fulcrum is at the center of gravity of the lever arm, This means that the lever arm itself creates no moment about the fulcrum because there is no moment arm for that particular force.

To solve, we sum moments about any convenient point to zero (zero because there is no acceleration in the F = ma equation)

The easiest convenient point is the fulcrum

30((90/2) - 15) - F(90/2) = 0

           30(30) = F(45)

                    F = 900/45 = 20 N

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Rank these significant figures numbers from the least to the most
Mumz [18]

Answer:

0.006<357<700.003<6010<9256.0<9520.00

8 0
3 years ago
Is cracking the eggs a physical change or a chemical change and why
matrenka [14]

Answer:

Cracking of an egg is a physical change since the egg and the stuff inside does not change but the shape or appearance of the shell changes.

Explanation:

Hope it helps

3 0
2 years ago
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where i
alukav5142 [94]

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

d=vt+\frac{1}{2}at^2

Now by substituting values in, we get

d=(0\frac{m}{s})*(2.01s)+\frac{1}{2}(9.8\frac{m}{s^2})(2.01)^2

d = 19.796m

7 0
3 years ago
A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc
Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

<u>Given the following data;</u>

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

<em>Angle = 41.7° Northeast.</em>

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

<em>Vr = 7.08m/s</em>

6 0
2 years ago
What is the average speed of a cheetah that runs 88 m in 5 seconds
vovangra [49]

So, the average speed of the Cheetah is 17.6 m/s.

<h3>Introduction</h3>

Hello ! I'm Deva from Brainly Indonesia. This time, I will help regarding the average speed. The average speed is obtained from finding the average of the speeds that occur or can be detected from the division between distance and travel time. The average speed can be formulated by :

\boxed{\sf{\bold{\overline{v} = \frac{s}{t}}}}

With the following condition :

  • \sf{\overline{v}} = average speed (m/s)
  • s = shift or distance objects from initial movement (m)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • s = shift = 88 m
  • t = interval of the time = 5 seconds

What was asked :

  • \sf{\overline{v}} = average speed = ... m/s

Step by step :

\sf{\overline{v} = \frac{s}{t}}

\sf{\overline{v} = \frac{88}{5}}

\boxed{\sf{\overline{v} = 17.6 \: m/s}}

So, the average speed of the Cheetah is 17.6 m/s.

5 0
2 years ago
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