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3 years ago

Which of the following is an illuminated object? a lightbulb a fire the Sun the Moon

2 answers:

7 0

The lightbulb, the fire, and the Sun all create light from some energy source.

The Moon doesn't create light. The parts of it that the Sun shines on
are visible, the parts of it that the Sun doesn't shine on are not visible.

Just like a rock on the ground at night, the Moon must be illuminated
by a flashlight in order to be seen. The Sun is the flashlight.

Neporo4naja [7]3 years ago

5 0

Answer:

LIGHTBULB

Explanation:

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a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(1

tekilochka [14]

By Newton's second law, the net force on the object is

∑ F = m a

∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N

Let f be the unknown force. Then

∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f

=> f = (-2.0 i - 12.0 j ) N

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3 years ago

What is the scientific definition of tracked?

dexar [7]

To find or discover by investigation?

8 0

3 years ago

Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

$-\tau_l+\tau_p+\tau_{cm}=0$ (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

$\tau=Fd$ (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

$F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N$

The force applied by the left support is 2123.33 N

8 0

3 years ago

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2

Harman [31]

When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.

Given:
a = 86 m/s^2
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m

vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m

Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m

5 0

3 years ago

Find the radioactivity of a 1 g sample of 226Ra given that <img src="https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D1620" id="TexFormul

dangina [55]

Answer:

Explanation:

No of atoms of Ra in 1 g of sample = 6.023 x 10²³ / 226

N = 2.66 x 10²¹

disintegration constant λ = .693 / half life

half life = 1620 x 365 x 60 x 60 x 24 = 5.1 x 10¹⁰ s

disintegration constant λ = .693 / 5.1 x 10¹⁰

radioactivity dn / dt = λN

= (.693 / 5.1 x 10¹⁰ ) x 2.66 x 10²¹

= .3614 x 10¹¹ per sec

= 3.614 x 10¹⁰ / s

6 0

3 years ago