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2 years ago

How much force is required to drag a 90 lb. box up this "frictionless" inclined plane? 109 lb. 10 lb. 81 lb. 9 lb.

Physics

2 answers:

MrRissso [65]2 years ago

5 0

I believe is 10 lb if not it's 9 lb.

Alex Ar [27]2 years ago

4 0

The correct answer is b) 10 lbs.

I hope that I was of help.

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If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit willa) decrease

slega [8]

Answer:

c) quadruple in magnitude

Explanation:

The power dissipated in the circuit is given by:

$P=I^2 R$

where

I is the current in the circuit

R is the total resistance of the circuit

In this problem:

- The current is doubled: I' = 2 I

- The resistance is kept constant: R' = R

So, the power dissipated is

$P' = (I')^2 R' = (2I)^2 R=4 I^2 R=4 P$

so, the power dissipated increase by a factor 4 (quadruples).

5 0

1 year ago

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

Anna35 [415]

This question is incomplete

Complete Question

Three equal point charges are held in place as shown in the figure below

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

A) F1=2F2

B) F1=3F2

C) F1=4F2

D) F1=9F2

Answer:

D) F1=9F2

Explanation:

We are told in the question that there are three equal point charges.

q, Q1, Q2 ,

q = Q1 = Q2

From the diagram we see the distance between the points d

q to Q1 = d

Q1 to Q2 = nd

Assuming n = 2

= 2 × d = 2d

Sum of the two distances = d + 2d = 3d

F1 is the force on q due to Q1 and

F2 is the force on q due to Q2,

Since we have 3 equal point charges and a total sum of distance which is 3d

Hence,

F1 = 9F2

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2 years ago

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liq [111]

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P = IV
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2 years ago

Which is a better conductor, a flagpole or a flag? Why?

Novosadov [1.4K]

Answer:

flagpole

Explanation:

if it is about electricity then its flagpole

4 0

1 year ago

Read 2 more answers

Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 14 × 15 × 8.0ft. the density of air

vova2212 [387]

Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

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Mass = (0.00118g/cm³)(0.0593 cm³)
Mass = 7 grams of HCN

7 0

1 year ago