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3 years ago

WILL REWARD 20 more pts once solved

Physics

1 answer:

Mandarinka [93]3 years ago

8 0

Explanation:

step 1. a diverging lens is "concave" on both side and always has a negative focal length

step 2. so 1/f = 1/s + 1/s' where f is the focal length, s is the object location, and s' is the image location (f, s, s' are all on the left side of the lens)

step 3. 1/-15 = 1/s + 1/-9 (image is virtual (negative))

step 4. 3/-45 = 1/s + 5/-45

step 5. s = 22.5cm (object is 22.5cm from lens)

step 6. s'/s = 9/22.5 ÷ 0.4 (magnification)

step 7. if the object is 4.5cm then the image is 4.5(0.4) = 1.8cm tall.

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A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find

garik1379 [7]

Refer to the figure shown below.

The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.

The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph

The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.

Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.

8 0

3 years ago

A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a

GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

$T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}$

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

$T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}$

T = 7.83 X10⁻⁷ s

6 0

3 years ago

A student is trying to determine the acceleration of a feather as she drops it to the ground. if the student is looking to achie

Anna [14]

The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.

All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.

If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.

5 0

3 years ago

Explain the difference between the three types of friction

USPshnik [31]

Static Friction

It is the friction that exists between a stationary object and the surface on which it's resting.

Sliding friction

It is the resistance created by two objects sliding against each other.

Rolling friction:-

It is the force resisting the motion when a body rolls on a surface.

hope this helps x

8 0

3 years ago

A diagram of a closed circuit with a power source on the left labeled 120 V. There are 3 resistors in parallel, separate paths,

Zina [86]

1) The equivalent resistance is $2.73\Omega$

2) The voltage in the circuit is 120 V

3) The total current in the circuit is 44.0 A

Explanation:

Resistors are said to be in parallel when they have the same potential difference at their terminals.

The formula to calculate the equivalent resistance for three resistors in parallel is:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

where $R_1, R_2, R_3$ are the resistances of the three resistors.

In this problem, the resistance of each resistor is:

$R_1 = 5.0 \Omega$

$R_2 = 10.0 \Omega$

$R_3 = 15.0 \Omega$

Substituting,

$\frac{1}{R_T}=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}=\frac{11}{30}$

So the equivalent resistance is

$R_T = \frac{30}{11}=2.73 \Omega$

In this problem we are told that the three resistors are connected in parallel. This means that their terminals are connected to the same point of the circuit: therefore, this means that they also have the same potential difference across them.

Therefore, the voltage in each branch containing each resistor is the same, and it is equal to the voltage of the battery, which is

$V=120 V$