The vector quantity that causes torsional acceleration of a rigid system is called ___

Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. h

Dvinal [7]

Answer:

734.16 kg m/ $s^{2}$

Explanation:

The problem is asking for the Force of pushing off the ground.

The formula of Force is: F = mass x acceleration

Given = Mass: 600 newtons (N)

Acceleration: 12 m/ $s^{2}$

We have to convert the mass into kg first. Remember that 1 kg is equal to 9.80665 newtons.

Let x be the mass in newtons.

Let's convert: $\frac{1 kg}{9.80665 N}$ x $\frac{x}{600 N}$ = $\frac{600}{9.80665}$ = 61.18 kg

Phil's weight is 61.18 kg

Let's go back to finding the force.

F = m x a

F = 61.18 kg x 12 m/ $s^{2}$

F = 734.16 kg m/ $s^{2}$

7 0

3 years ago

How are watts measured?

BARSIC [14]

Answer:

The watt (symbol: W) is a unit of power. In the International System of Units (SI) it is defined as a derived unit of 1 joule per second, and is used to quantify the rate of energy transfer

Explanation:

8 0

3 years ago

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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

A 0.5 kg block is attached to a spring (k = 12.5 N/m). The damped frequency is 0.2% lower than the natural frequency, (a) What i

Tju [1.3M]

Answer:

a) 1.58 kg s^{-1}

b) x_m e^{-1.58t} x_m is initial amplitude

c) 5 kg s^{-1}

Explanation:

given data:

mass =0.5 kg

k = 12.5 N/m

from the data given

a) $w_d = w_o - \frac{0.2}{100}w_o$

$= w_o - 0.002w_o = 0.99w_o$

$w_d = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}$

$0.998w_o = \sqrt{w_o^2 - \frac{b^2}{4m^2}$

$(0.998w_o)^2 = w_o^2 -\frac{b^2}{1}$

$b^2 = w_o^2 -(0.998w_o)^2$

$b^2 = w_o^2(1-0.998^2) = 3.996 *10^{-3} w_o^2$

$b = w_o\sqrt{3.996*10^{-3}}$

$b = \frac{12.5}{0.5}\sqrt{3.996*10^{-3}} = 1.58 kg s^{-1}$

b) $x = x_m e^{\frac{-bt}{2m}}$

$x = x_m e^{-1.58t}{1} = x_m e^{-1.58t}$ where x_m is initial amplitude

c) critical damping amplitude

$c_c =2\sqrt{km} = 2\sqrt{12.5*.5} = 5 kg s^{-1}$

5 0

3 years ago

Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

EleoNora [17]

3) Earth is about 150 million km from the Sun, and the apparent brightness of the Sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the Sun if we were located five times Earth's distance from the Sun. Answer: The Sun would appear 1/25 times as bright.

4 0

4 years ago

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The vector quantity that causes torsional acceleration of a rigid system is called ____