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djverab [1.8K]
3 years ago
5

A crate is sliding down an inclined ramp at a constant speed of 0.55 m/s. The vector sum of all the forces acting on this crate

must point down the ramp
Physics
1 answer:
Lemur [1.5K]3 years ago
8 0

Answer:

F_{net} = 0

Explanation:

here it is given that the crate is sliding down with constant speed

so we will have

v = constant

now we can say

a = \frac{dv}{dt}

so here we have

a = 0

now from Newton's II law we can say that sum of all forces on the system is always product of mass and its acceleration

so here we will have

F_{net} = ma

F_{net} = 0

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Answer:

yes

Explanation:

I think so because it not mention in the law

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The three main types of nuclear radiation are alpha, beta, and gamma
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Gamma beta alpha

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2. For each of the listed parts of a power plant, make a selection to indicate in what
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Answer: Find the answer in the explanation

Explanation: Given the Roman numeral and the representation

I. part of a coal-fired power plant

II. part of a nuclear power plant

III. part of a coal-fired power plant and part of a nuclear power plant

a.) Boiler : I

b.) Combustion chamber: I

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3 years ago
At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th
Korvikt [17]

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

8 0
3 years ago
A wire of radius 5 x 10⁻⁴ m is needed to prepare a coil of resistance 40 Ω. The resistivity of the material of the wire is 3.14x
SIZIF [17.4K]

Answer:

100 m

Explanation:

From the question,

R = Lρ/A.................... Equation 1

Where R = resistance of the wire, L = length of the wire, ρ = resistivity of the wire, A = cross sectional area of the wire.

But,

A = πr².................... Equation 2

Where r = radius of the wire.

Substitute equation 2 into equation 1

R = Lρ/πr²

Make L the subject of the equation

L = Rπr²/ρ...................... Equation 3

Given: R = 40 Ω, r = 5×10⁻⁴ m, ρ = 3.14×10⁻⁷ Ωm

Constant: π = 3.14

Substitute these values into equation 3

L = [40×3.14×( 5×10⁻⁴)²]/ (3.14×10⁻⁷)

L = 40×3.14×25×10⁻⁸/(3.14×10⁻⁷)

L = 100 m

Hence the length of the wire is 100 m

4 0
2 years ago
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