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2 years ago

Help plzzzzzzzzzzzzzzzz!

Physics

1 answer:

aksik [14]2 years ago

3 0

concave <span>ray diagrams were constructed in order to determine the general location, size, orientation, and type of image formed by concave mirrors. Perhaps you noticed that there is a definite relationship between the image characteristics and the location where an object placed in front of a concave mirror. but, convex</span><span>ray diagrams were constructed in order to determine the location, size, orientation, and type of image formed by concave mirrors. The ray diagram constructed earlier for a convex mirror revealed that the image of the object was virtual, upright, reduced in size and located behind the mirror. </span>

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A spring has a spring constant of 100 N/m, and the mass hanging from is is 0.71 kg. What is the period of the spring's motion?

Anuta_ua [19.1K]

k = spring constant of the spring = 100 N/m

m = mass hanging from the spring = 0.71 kg

T = Time period of the spring's motion = ?

Time period of the oscillations of the mass hanging is given as

T = (2π) √(m/k)

inserting the values in the above equation

T = (2 x 3.14) √(0.71 kg/100 N/m)

T = (6.28) √(0.0071 sec²)

T = (6.28) (0.084) sec

T = 0.53 sec

hence the correct choice is D) 0.53

6 0

3 years ago

What is the net force acting on a .15 kg hockey puck accelerating at a rate of 12 m/s2

maw [93]

Answer:

The net force is 1.8N

Explanation:

Given that the formula for force is Force = mass×acceleration. So you have to substitute the values into the formula :

$force = mass \times acceleration$

Let mass = 0.15kg,

Let acceleration = 12m/s²,

$force = 0.15 \times 12$

$force = 1.8$

7 0

3 years ago

Read 2 more answers

An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units.

makkiz [27]

Answer:

Acceleration of the object is $4\ m/s^2$ .

Explanation:

It is given that, the position of the object is given by :

$r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j$

Velocity of the object, $v=\dfrac{dr}{dt}$

Acceleration of the object is given by :

$a=\dfrac{d^2r}{dt^2}$

$a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)$

Using the property of differentiation, we get :

$a=\dfrac{d^2r}{dt^2}=-4\ m/s^2$

So, the magnitude of the acceleration of the object at time t = 2.00 s is $4\ m/s^2$ . Hence, this is the required solution.

5 0

2 years ago

What is the gravitational potential energy of a 150 kg object suspended 5m above the Earth's surface

ella [17]

The gravitational potential energy referred to the ground level is given by
$U=mgh$
where m is the mass of the object, $g=9.81 m/s^2$ is the gravitational acceleration and h is the height of the object with respect to the ground.

Therefore in our problem the potential energy is
$U=(150 kg)(9.81 m/s^2)(5 m)=7357.5 J$

6 0

3 years ago

The chart shows the temperatures of four different substances.

garik1379 [7]

Answer:

C) 20

Explanation:

Happy Halloween LOL

4 0

3 years ago

Read 2 more answers