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OverLord2011 [107]

2 years ago

6

If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan

t velocity? a no force applied b a greater force c an equal force d a lesser force

1 answer:

nlexa [21]2 years ago

5 0

Answer:

Explanation:

hi

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aleksley [76]

The meters per second
+1t a second / 2t

6 0

2 years ago

A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the

pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²

W(1 to 2) = 11.25 J

W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0 to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

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6 0

1 year ago

What is the MAIN cause of increased erosion, especially for soil?

egoroff_w [7]

I think it’s C. But I’m not completely sure.

7 0

2 years ago

Three capacitors (4.7, 6.5, and 15.5 F) are connected in series across a 50.0-V battery. Find the voltage across the 4.7-F capac

satela [25.4K]

Answer: 50 V

Explanation:

Given

Three capacitors are connected in parallel across 50 V battery

When capacitor are connected in parallel, net capacitance is the sum of all the values

Net capacitance

$\Rightarrow C=4.7+6.5+15.5\\\Rightarrow C=26.7\ F$

Voltage in parallel circuit is same across each capacitor. Therefore, voltage across 4.7 F capacitor is 50 V

6 0

2 years ago

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,

KATRIN_1 [288]

Answer:

a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

V = i (R₁ + R₂)

with R₁ connected

V = (i + 0.5) R₁

with R₂ connected

V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

V = i ( $\frac{V}{ i+0.5} + \frac{V}{i+0.25}$ )

1 = i ( $\frac{1}{i+0.5} + \frac{1}{i+0.25}$ )

1 = i ( $\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }$ ) = $\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}$

i² + 0.75 i + 0.125 = 2i² + 0.75 i

i² - 0.125 = 0

i = √0.125

i = 0.35355 A

with the second equation we look for R1

R₁ = $\frac{V}{i+0.5}$

R₁ = 12 /( 0.35355 +0.5)

R₁ = 14.1 Ω

with the third equation we look for R2

R₂ = $\frac{V}{i+0.25}$

R₂ = $\frac{12}{0.35355+0.25}$

R₂ = 19.9 Ω

3 0

2 years ago