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myrzilka [38]
3 years ago
6

I need to know the answer with steps plz

Physics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

a.) 40cm^3

b.) 5 blocks

Explanation:

Let's find the volume of the metal block:

V=lwh

V=(10cm)(2cm)(2cm)

V=40cm^3

Now, let's find the volume of the smaller blocks:

V=lwh

V=(2cm)(2cm)(2cm)

V=8cm^3

So, one small block has a volume of 8cm. If the volume of the larger block is 40cm, we can fit a total of 5 smaller blocks into the larger block, meaning it will make up the same volume.

\frac{40}{8} = 5

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An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

7 0
4 years ago
Given uranium 235 go through an alpha, beta, beta, alpha, gamma, neutron and alpha. What do you have now?
Licemer1 [7]

Answer:

b. 88, 222

Explanation:

235U₉₂ ----→ Alpha --------→  231P₉₀  ----→- beta -----→ 231Q₉₁ ------→-beta -------→231R₉₂--------→-alpha ------→-227S₉₀ ------→ gamma -----→-227S₉₀ ----------→ neutron ------→-226T₉₀-----------→ alpha --------→222 X ₈₈

Atomic No is 88 , atomic weight = 222 .

3 0
3 years ago
The first law of thermodynamics is just another form of the _____.
Oksi-84 [34.3K]
A. Conservation of energy
8 0
3 years ago
Why does the weight of water pulls the central part of the surface down?
Brut [27]

Answer:

Buoyancy or Upthrust

Explanation:

This is an upward force exerted by a fluid that opposes the weight of a partially or fully immersed object.

5 0
3 years ago
9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal
natima [27]

Answer:

a. ω₂ = 14rad/s

b. ∇K.E = 0.014J

c. The bug does not conserve force while moving on the disk (non-conservative force).

Explanation:

Mass of the bug (m) = 0.02kg

Mass of the cylindrical disk (M) = 0.10kg

Radius of the disk (r) = 0.10m

Initial angular velocity ω₁ = 10rad/s

final angular velocity ω₂ = ?

a.

To calculate the new angular velocity, we relate it to the conservation of angular momentum of the system I.e when the bug was at the edge of the disk and when it is located at the centre of the disk.

I = Mr² / 2

I₁ = Mr₂ / 2 + mr²

I₁ = moment of inertia when the bug was at the edge

I₁ = [(0.10 * 0.10²) / 2 ] + (0.02 * 0.1²)

I₁ = 0.0005 + 0.0002

I₁ = 7.0*10⁻⁴kgm²

I₂ = moment of inertia when yhe bug was at the center of the disk.

I₂ = Mr² / 2

I₂ = (0.01 * 0.01²) 2

I₂ = 0.0005kgm²

for conservation of angular momentum,

I₁ω₁ = I₂ω₂

solve for ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (7.0*10⁻⁴ * 10) / 5.0*10⁻⁴

ω₂ = 14 rad/s

b. the change in kinetic energy of the system is

∇K = K₂ - K₁

∇K = ½I₂*ω₂² - ½I₁*ω₁²

∇K = ½(I₂*ω₂² - I₁ω₁²)

∇K = ½[(5.0*10⁻⁴ * 14²) - (7.0*10⁻⁴ * 10²)]

∇k = ½(0.098 - 0.07)

∇K = ½ * 0.028

∇K = 0.014J

c.

The cause of the decrease and increase in kinetic energy is because the bug uses a non-conservative force. To conserve the mechanical energy of a system, all the forces acting in it must be conservative.

The work W produced by this force brings the difference in kinetic energy of the system

W = K₂ - K₁

6 0
3 years ago
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