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3 years ago

Please help me with both of them

Physics

1 answer:

8_murik_8 [283]3 years ago

4 0

Answer:

101011010101010100101000101010100010011010100010100000101041204105210241012021012021021012022221222122122345788981633333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333311111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111114565656565+4+652147

21212121512

546213171549895465621324547998995656565656565656565722426579898541321447985331321

Explanation:

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A worker pushed a 33 kg block 6.1 m along a level floor at constant speed with a force directed 23° below the horizontal. if the

jenyasd209 [6]

The work done occurs only in the direction the block was moved - horizontally. Work is given by:

W = F(h) * d

Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.

Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):

F(h) = F(app)cos(23)

Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:

N = mg + F(v) = mg + F(app)sin(23)

Now we can get down to business and solve for F(app) - as mentioned above:

F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8

Now that we have F(app), we can find the exact value of F(h):

F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7

And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3

Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.

3 0

3 years ago

1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into t

AnnZ [28]

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, $V_T$ , is given by the sum of the rock, $V_r$ , and the water, V, is given as follows;

$V_T$ = $V_r$ + V

$V_T$ = A × h₂

∴ $V_T$ = 10 cm² × 12 cm = 120 cm³

The total volume, $V_T$ = 120 cm³

The volume of the rock, $V_r$ = $V_T$ - V

∴ $V_r$ = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, $V_r$ = 20 cm³

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

8 0

3 years ago

Init.

nikklg [1K]

Answer:

The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.

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