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zhuklara [117]
3 years ago
5

The fulcrum of a uniform 20-kg seesaw that is 4.0 m long is located 2.5 m from one end. A 30-kg child sits on the long end. Part

A Determine the mass a person at the other end would have to be in order to balance the seesaw. Express your answer to two significant figures and include the appropriate units. m = nothing nothing

Physics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

57 kg

Explanation:

Mass of seesaw = 20 kg

Length of seesaw = 4 m

Mass of child on the longer end = 30 kg

The weight of the seesaw acts at the center i.e. 2m

The algebraic sum of moments of all forces about any point is zero, hence, using the fulcrum as the reference point:

[x * 9.8* 1.5] - [20 * 9.8* (2.5 - 2)] - [30 * 9.8 * 2.5] = 0

=> 14.7x = (20*9.8*0.5) + 735

14.7x = 98 + 735

14.7x = 833

=> x = 833/14.7

x = 57 kg

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David is driving a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady
SpyIntel [72]

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.

7 0
2 years ago
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A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from
ladessa [460]

Answer:

P_1 = 1166.7 Watt

P_2 = 2000 Watt

Explanation:

Average power for the human sprinter is given as

Power = \frac{\Delta E}{\Delta t}

so we have

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}

P_1 = 1166.7 Watt

Average power for greyhound is given as

P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}

P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}

P_2 = 2000 Watt

3 0
3 years ago
Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the
mash [69]

Answer:

The value is }  N  =  66 \  blocks

Explanation:

From the question we are told that

The weight of the block is W_b  = 100 \  N

The dimension of the block is d =  0.400 m  \ X  \ 0.250 \  m  \  X  \ 0.130 \ m

Generally two atmosphere is equivalent to

P_{2atm} =  2 *  1.013 *10^{5} =  202600 \  N/m^2

Generally 1 atm = 1.013 *10^{5} N/m^2

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

So

A =  0.250 *  0.130

=> A =  0.0325 \  m^2

Generally the force due to this blocks is mathematically represented as

F =  N  *  W_b

Here N is the number of blocks

So

}  202600 =  \frac{N  *  100 }{ 0.0325}

=>   }  N  =  66 \  blocks

3 0
3 years ago
30 points!
vlada-n [284]

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc)  items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

6 0
2 years ago
a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the
Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 0
2 years ago
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