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Ymorist [56]
3 years ago
13

Which of the following is formed by constructive erosion? 1. delta 2. gully 3. rill 4. glacier

Physics
1 answer:
lana [24]3 years ago
7 0
1. Delta, is formed by constructive erosion.
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Help me with this please !!
Phoenix [80]
25.4is the right answer
4 0
3 years ago
Read 2 more answers
Student pushes a 50 N block across the floor for a distance of 15 m how much work was done to move the block
Talja [164]

Answer:

750 J

Explanation:

We have a student that pushes a 50N block  across the floor for a distance of 15m. The question is asking how much work was done to move the block.

To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS

F being the force and S being the distance.

W = FS

W = (50)(15)

W = 750

Therefore, 750 joules is our answer.

7 0
3 years ago
Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same len
Mekhanik [1.2K]

Answer:

  • The difference in length for steel is 2.46 x 10⁻⁴ m
  • The difference in length for invar is 1.845 x 10⁻⁵ m

Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, \gamma_{st.} =  3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, \gamma_{in.} =  2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂  = L₁ + L₁αθ

L₂  - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = \frac{\gamma}{3}

Difference in length, for steel at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC

ΔL  = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL =  L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC

ΔL  = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

8 0
3 years ago
Does anyone know how to solve this? 4+9×2÷3−1
Anika [276]

The answer is 7.66. c:

6 0
3 years ago
Read 2 more answers
Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you
Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

(c) t = 27.2 min

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light  λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

      = 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.  

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

b)

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A =  6.14 * 10⁻⁴ / 1.0 * 10⁶ =  6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

c)  

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t  

v = u+at  

t = 1.0 - 0/ 6.14 * 10⁻⁴ =  1629s = 27.2 min

t = 27.2 min

4 0
4 years ago
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