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Julli [10]
3 years ago
6

Tim jogs a distance of 7.2 km to the west. Then he turns south and jogs 1.4 km. West is the resultant if Tim's jog back to the b

eginning?
Physics
1 answer:
vesna_86 [32]3 years ago
3 0

Answer:

Explanation:

If Tim jogs a distance of 7.2 km to the west and then he turns south and jogs 1.4 km, the resultant displacement of Tim is calculated using the pythagoras theorem as shown;

R² = 7.2²+1.4²

R² = 51.84+1.96

R² = 53.8

R = √53.8

R = 7.33 km

Hence  the resultant of Tim's jog back to the beginning is 7.33km

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Density is mass over volume. Density is also a charecteristic property wich defines a substance and two different substance cant have the same charasteristic property.
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Matthew is waterskiing. As the boat starts moving, he is at an angle of 8.0° to the right of the boat. The boat applies 250 newt
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The work done is B. 1.2\cdot 10^4 J

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the boat in this problem, we have:

F = 250 N (force applied)

d = 50 m (displacement)

\theta=8.0^{\circ}

Substituting, we find the work done:

W=(250)(50)(cos 8^{\circ})=1.2\cdot 10^4 J

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If a car has a mass of 1,000 kilograms, and a velocity of 35 m/s. What is its momentum?
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A beam of electrons with of wavelength of 7.5 x 10-6 m is incident on a pair of narrow rectangular slits separated by 0.75 mm. T
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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
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