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3 years ago

A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?

Physics

1 answer:

Eduardwww [97]3 years ago

6 0

relation between linear velocity and angular velocity is given as

$v = R\omega$

here

v = linear speed

R = radius

$\omega$ = angular speed

now plug in all data in the equation

$25.0 = 0.410 \omega$

$\omega = \frac{25}{0.410}$

$\omega = 60.9 rad/s$

so rotating speed is 60.9 rad/s

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A box full of stuff was pushed for 500 N at 15 m/s2. What is the mass of the box?

stepladder [879]

F = ma
500 = m x 15
m = 33.33 kg

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3 years ago

A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d

andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x² - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0

3 years ago

The relationship that exists between gravity and distance and mass respectively.

ruslelena [56]

Answer:

D...............................

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3 years ago

4. A bullet of mass 30 g is fired from a rifle of mass 5kg at a speed of 259m/s.

Ostrovityanka [42]

Answer:

Rifle Momentum=7.77kg*m/s v'= 1.554 m/s

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

0+0 = 0.03*259 + P(rifle momentum)

solve for P

p= 7.77kg*m/s

b) 7.77= 5*v'

v'= 1.554 m/s

8 0

3 years ago

A car covers 72 kilometers in the first hour of its journey. In the next hour, it covers 90 kilometers. What is the amount of wo

Nina [5.8K]

The work done is $2.8125 \times 10^{5} \mathrm{J}$

Work Done = Change in Kinetic Energy (ΔKE)

<u>Explanation</u>

In first 1 hour it travels 72 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{72}{1} k m / h=72 \mathrm{km} / \mathrm{h}=\frac{72000}{3600} \mathrm{m} / \mathrm{s}=20 \mathrm{m} / \mathrm{s}$

or, Initial Velocity (u) = 20 m/s

Similarly for the next hour it covers 90 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{90}{1} k m / h=90 \mathrm{km} / \mathrm{h}=\frac{90000}{3600} \mathrm{m} / \mathrm{s}=25 \mathrm{m} / \mathrm{s}$

or, Final Velocity (v) = 20 m/s

Work done = Change in Kinetic Energy (ΔKE)

Work done = ΔKE = $\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$

ΔKE = $\frac{1}{2} m\left(v^{2}-u^{2}\right)=\frac{1}{2} \times\left(2.5 \times 10^{3}\right) \times\left(25^{2}-20^{2}\right)$

= $\frac{2500 \times(625-400)}{2}=\frac{2500 \times 225}{2}=\frac{562500}{2}$ = 281250 joule

= $2.8125 \times 10^{5} \mathrm{J}$

4 0

3 years ago