It would be 2300000 grams.
Answer:
Resistance =330 Ω
Tolerance = 33 Ω
Explanation:
see attached resistor color code table
The first stripe is orange, which means the leftmost digit is a 3.
The second stripe is orange , which means the next digit is a 3.
The third stripe is brown. Since brown is 1, it means add one zero to the right of the first two digits.
The resistance is:
orange-orange-brown= 330 Ω
The tolerance is:
The fourth color band indicates the resistor's tolerance. Tolerance is the percentage of error in the resistor's resistance.
silver is 10%
A 330 Ω resistor has a silver tolerance band.
<em>Tolerance = value of resistor x value of tolerance band </em>
= 330 Ω x 10% = 33 Ω
330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.
Answer:
15.7 m
Explanation:
The range (horizontal distance) of the projectile is determined only by its horizontal motion.
The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

where
v = 12.0 m/s is the initial velocity
is the angle between the direction of v and the horizontal
Substituting,

We know that the projectile hits the ground in a time of
t = 2.08 s
so the horizontal distance covered is

Answer:
the force will increase by a factor 2.25
Explanation:
The gravitational force between the two stars is given by:

where
G is the gravitational constant
m1, m2 are the masses of the two stars
r is the distance between the stars
If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

So the new force will be

So, the force will be 2.25 times the previous value.
I think you almost got it.
At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.
With the value of the initial speed (28 m/s, which is the total speed), you can set
v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7
or theta = 64.62 deg, it is D. Think about it. I hope you see it.