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2 years ago

A force of 88 N is used to move a couch 2.1 m. How much work is done?

1 answer:

7 0

$Given:\\F=88N\\s=2.1m\\\\Find:\\W=?\\\\Solution:\\\\W=\overrightarrow{F}\circ\overrightarrow{s}\\\\In\;these\;case\\\\W=Fs\;\;\;\;because\;\angle(\overrightarrow{F},\overrightarrow{s})=0\\\\W=88N\cdot 2.1m=184.8J\\\\Correct\;answer\;is\;A$

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Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x

never [62]

Answer:

210.3 degrees

Explanation:

The net force exerted on charge A = 59.5 N

Use the x and y coordinates of net force to get the direction

arctan (y/x)

8 0

2 years ago

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17. The edges of a rectangular solid have these measures: 1.5 feet by 1½ feet by 3 inches. What is its volume in cubic inches? a

Alisiya [41]

Answer:

c. 972

Explanation:

The volume of a rectangular solid is calculated as the product of its dimensions, that is, its width, its length and its height:

$V=a*b*h$

1 feet is equal to 12 inches, so:

$a=1.5ft*\frac{12in}{1ft}=18in\\b=(1+\frac{1}{2})ft\\b=\frac{3}{2}ft*\frac{12in}{1ft}=18in$

Now, we calculate the volume of the object in cubic inches:

$V=18in*18in*3in\\V=972in^3$

6 0

2 years ago

A student decides to give his bicycle a tune up. He flips it upside down (so there’s no friction with the ground) and applies a

Alinara [238K]

Answer:

Tangential velocity = 10.9 m/S

Explanation:

As per the data given in the question,

Force = 20 N

Time = 1.2 S

Length = 16.5 cm

Radius = 33.0 cm

Moment of inertia = 1200 kg.cm^2 = 1200 × 10^(-4) kg.m^2

= 1200 × 10^(-2) m^2

Revolution of the pedal ÷ revolution of wheel = 1

Torque on the pedal = Force × Length

= 20 × 16.5 10^(-2)

= 3.30 N m

So, Angular acceleration = Torque ÷ Moment of inertia

= 3.30 ÷ 12 × 10^(-2)

= 27.50 rad ÷ S^2

Since wheel started rotating from rest, so initial angular velocity = 0 rad/S

Now, Angular velocity = Initial angular velocity + Angular Acceleration × Time

= 0 + 27.50 × 1.2

= 33 rad/S

Hence, Tangential velocity = Angular velocity × Radius

= 33 × 33 × 10^(-2)

= 10.9 m/S

7 0

2 years ago

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

kipiarov [429]

Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

7 0

2 years ago

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A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe

natta225 [31]

First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.

Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.

4 0

2 years ago