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oksian1 [2.3K]
3 years ago
10

A block of metal of density 3000 kg/m3 is 2m high and stands on a square base of side 0.5 m. Calculate:

Physics
1 answer:
labwork [276]3 years ago
8 0

Answer:

1)  A = 0.25 m², 2) V = 0.5 m³, 3)   m = 1500 kg, 4) W = 14700 N,

5)  P = 58800 Pa

Explanation:

1) The area of ​​the base is square

          A = L²

         A = 0.5²

         A = 0.25 m²

2) The block is a parallelepiped

         V = A h

         V = 0.25 2

          V = 0.5 m³

3) Density is defined

           rho = m / V

           m = rho V

           m = 3000 0.5

           m = 1500 kg

4) The weight of a body is

           W = mg

            W = 1500 9.8

            W = 14700 N

5) The pressure is

             P = F / A

in this case the force is equal to the weight of the body

              P = 14700 / 0.25

              P = 58800 Pa

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7 0
2 years ago
A 103 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 93 m below
siniylev [52]

Answer:

125.83672 seconds

Explanation:

P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)

m = Mass of professor = 103 kg

g = Acceleration due to gravity = 9.8 m/s²

h = Height of professor = 93 m

Work done would be equal to the potential energy

W=mgh\\\Rightarrow W=103\times 9.8\times 93\\\Rightarrow W=93874.2\ J

Power is given by

P=\frac{W}{t}\\\Rightarrow t=\frac{W}{P}\\\Rightarrow t=\frac{93874.2}{746}\\\Rightarrow t=125.83672\ seconds

The time taken by the horse to pull the professor is 125.83672 seconds

6 0
3 years ago
A sprinter runs for 6.45 s at 7.44 m/s. How far does she get?
Gwar [14]

Answer:

D is the answer

Explanation:

6.45×7.44= 47.98800

Which if we round of we get 48m

8 0
3 years ago
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Answer:

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3 0
3 years ago
A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to
Arisa [49]

Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

3 0
3 years ago
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