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xenn [34]
3 years ago
8

The diagram below shows a flat mirror reflecting

Physics
2 answers:
egoroff_w [7]3 years ago
7 0

Answer:

c

Explanation

your welcome

sukhopar [10]3 years ago
6 0

Answer:

TO MEASURE THE ANGLES OF RAYS

Explanation:

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How much force is required to accelerate a 2kg mass at 3 m/s2? answer key?
mr Goodwill [35]
F = ma so u can plug in the given numbers and solve:

F = (2)(3)
4 0
3 years ago
A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i
Lubov Fominskaja [6]

Answer:

11.8 m/s

Explanation:

At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).

Sum of forces in the centripetal direction:

∑F = ma

mg − N = m v²/r

At the maximum speed, the normal force is 0.

mg = m v²/r

g = v²/r

v = √(gr)

v = √(9.8 m/s² × 14.2 m)

v = 11.8 m/s

3 0
2 years ago
An object traveling at a constant speed but with a changing direction is accelerating.
prohojiy [21]

Strange as it may seem, that's true. (choice 'a'.)

"Acceleration" doesn't mean "speeding up".  It means ANY change in
the speed or direction of motion.  So a car with the brakes applied
and slowing down, and a point on the rim of a bicycle wheel that's
turning at a constant rate, are both accelerating.

6 0
2 years ago
Read 2 more answers
Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth
USPshnik [31]

The time needed for the hammer to reach the surface of the Earth is 3.54 s.

<h3>Time of motion of the hammer</h3>

The time of motion is calculated as follows;

t = √(2h/g)

where;

  • h is height of fall
  • g is acceleration due to gravity

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

8 0
2 years ago
A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside
RUDIKE [14]

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

B = \frac{\mu_o NI}{l} \\\\

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

8 0
3 years ago
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