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yan [13]
3 years ago
10

Find the magnitude of the torque produced by a 4.5 N force applied to a door at a perpendicular distance of 0.26 m from the hing

e. Answer in units of N · m.
Physics
1 answer:
Alex787 [66]3 years ago
3 0

Answer:

The required torque is 1.17 N-m.

Explanation:

The given data :-

The magnitude of force ( F ) = 4.5 N.

The length of arm ( r ) = 0.26 m.

Here given that force is applied at perpendicular means ( ∅ ) = 90°.

The torque ( T ) is given by

T = F * r * sin∅

T = 4.5 * 0.26 * sin 90°

T = 4.5 * 0.26 * 1

T = 1.17 N-m.

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A vector has components Ax = 12.0 m and Ay= 5.00 m. What is the angle that vector A makes with the x-axis?a. 67.4ob. 32.6oc. 22.
antoniya [11.8K]

Answer:

C.22.6°

Explanation:

Ax = ACosθ,

A = Ax/Cosθ    .....equ1

Ay = Asinθ,

A= Ay/sinθ      .....equ 2

equate equation 1 and 2.

Ax/Cosθ = Ay/sinθ

Ax = 12.0 m,  Ay= 5.00 m

12/Cosθ = 5/sinθ

θ = 22.6°

5 0
3 years ago
6
choli [55]

Answer:

B. Acceleration in the direction of motion speeds you up

Explanation:

Acceleration is defined as when something gains speed. For example, when a car speeds up. Deceleration is when the car slows down, and looses speed. When defining these terms, think of a car going faster, then slowing down at a red light.

3 0
3 years ago
Suppose a radar gun sends out radio waves with a frequency of 2,000,000.0 Hz. The waves bounce off a moving car and return with
Igoryamba

Answer:

300,000,030m/s

Explanation:

Using the relationship between frequency and speed of a wave;

F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k

Let F1 be the frequency of the radio wave = 2,000,000.0Hz

V1 be the speed of light = 300,000,000m/s

F2 = frequency produced by the car = 2,000,000.2Hz

V2 be the velocity of the moving car

Substituting this values in the equation above;

2,000,000/300,000,000 = 2,000,000.2/V2

Cross multiplying

2,000,000V2 = 300,000,000×2,000,000.2

2V2 = 300×2,000,000.2

V2 = 150×2,000,000.2

V2 = 300,000,030m/s

The velocity of the car is 300,000,030m/s

6 0
3 years ago
Fill in the blanks for the following:
storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm-mol^{-1}K^{-1}

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = \sqrt{\frac{3RT}{M} }

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = \sqrt{\frac{3*8.314*293}{0.0319} }= <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

\frac{Voxy}{Vnit} = \sqrt{\frac{Mnit}{Moxy} }

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

\frac{478.6}{Vnit} = \sqrt{\frac{14.0}{31.9} }

\frac{478.6}{Vnit} = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

6 0
2 years ago
A rug sits in a sunny place on the floor for a long time. Kiara thinks that light from the sun can cause the rug's color to fade
geniusboy [140]

Answer:

Yes! Light from the sun can affect the materials certain carpets are made out of. The usual effect being the dye in the carpet being "washed out" or "dried out" as the sun beams down on it. When this happens, the carpet will usually lose its color, causing it to fade.

3 0
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