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Novosadov [1.4K]
3 years ago
8

A small block of mass 1.57 kg rests on the left edge of a block of length 3.45 m and mass 7.04 kg. the coefficient of kinetic fr

iction between the two blocks is 0.307, and the surface on which the 7.04 kg block rests is frictionless. a constant horizontal force of magnitude 10.4 n is applied to the 1.57 kg block, setting it in motion. 3.45 m 10.4 n 7.04 kg 1.57 kg 1.57 kg 7.04 kg 10.4 n how long will it take before this block reaches the right side of the 7.04 kg block? the acceleration of gravity is 9.8 m/s 2 . (note that both blocks are set in motion when f~ is applied. answer in units of s. 002 (part 2 of 2) 10.0 points how far did the 7.04 kg block move in
Physics
1 answer:
MariettaO [177]3 years ago
3 0
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3. A 900N mountain climber scales a
Umnica [9.8K]

Answer:

<h2>135,000 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 900 × 150

We have the final answer as

<h3>135,000 J</h3>

Hope this helps you

4 0
3 years ago
How might the velocity (speed) of wind or water affect the deposition of sediments?
Maksim231197 [3]
Well depending on the speed of both of those things is were the rock will be placed and it also determines how fast can an environment change
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6 0
4 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i
Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.  

5 0
3 years ago
When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl
zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

If

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L =  ρ*I / S

then

(V / L)₁ = (V / L)₂  ⇒     ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

If

S₁ = 0.25*π*D²   and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒    I₂ = 4*ρ₁*I₁ / ρ₂

⇒     I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒     I₂ = 21.13 mA

5 0
3 years ago
Help me with this question, please
Lunna [17]

Answer:

First one is voltage

second is resistance

third is electric current

hope this helped!!

5 0
3 years ago
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