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damaskus [11]
3 years ago
6

What are some examples for number 4?

Physics
1 answer:
lidiya [134]3 years ago
5 0

I would love to help you, but the picture is in a difficult position to read.

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A concave mirror has a focal length of 30.0 CM. an object is placed 15.0 CM from the mirror. what is the radius of curvature of
Nana76 [90]

Answer:

60 cm

Explanation:

We are given;

  • Focal length of a concave mirror as  30.0 cm
  • Object distance is 15.0 cm

We are required to determine the radius of curvature.

We need to know that the radius of a curvature is the radius of a circle from which the curved mirror is part.

We also need to know that the radius of curvature is twice the focal length of a curved mirror.

Therefore;

Radius of curvature = 2 × Focal length

Therefore;

Radius of curvature = 2 × 30 cm

                                 = 60 cm                                  

7 0
3 years ago
The sun rose at 6:54 this morning.<br>is this an observation or inference?
jeka57 [31]


Observation: You saw/ observed something and you are stating that.

Inference: You are predicting something to be true. If so and so, then this will occur.

4 0
3 years ago
Read 2 more answers
h(t) = - 16t2 + 64t + 112 where t is the time in seconds. After how many seconds does the arrow reach it maximum height? Round t
laila [671]

Answer:

2 seconds

Explanation:

The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.

h(t)=-16t^{2}+64t+112

Differentiate with respect to t on both the sides, we get

\frac{dh}{dt}=-32t+64

For maxima and minima, put the value of dh / dt is equal to zero. we get

- 32 t + 64 = 0

t = 2 second

Thus, the arrow reaches at maximum height after 2 seconds.

8 0
3 years ago
A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the
guapka [62]

Answer:

K.E =  \frac{1}{2} m {v}^{2}  \\  {v}^{2}_i  =  {v}^{2} _x + {v}^{2} _y \\  =  {(6)}^{2}  +  {( - 2)}^{2}  = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (40) = 60J \\  \\ {v}^{2}_f  =  {v}^{2} _x + {v}^{2} _y \\  =  {(8)}^{2}  +  {(4)}^{2}  = 80m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\  = 120J - 60J \\  = 60J

3 0
3 years ago
The gun, mount, and train car of a railway had a total mass of 1.22 x 10^6 kg. The gun fired a projectile that was 80 cm in diam
kakasveta [241]
1) According to the law of conservation of momentum .. 
<span>Horiz recoil mom of gun (M x v) = horiz. mon acquired by shell (m x Vh) </span>

<span>1.22^6kg x 5.0 m/s = 7502kg x Vh </span>
<span>Vh = 1.22^6 x 5 / 7502 .. .. Vh = 813 m/s </span>

<span>Barrel velocity V .. .. cos20 = Vh / V .. ..V = 813 /cos20 .. .. ►V = 865 m/s </span>

<span>2) Using the standard range equation .. R = u² sin2θ /g </span>

<span>R = 865² x sin40 / 9.80 .. .. ►R = 49077 m .. (49 km)</span>
5 0
3 years ago
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