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stepladder [879]
3 years ago
13

1. The current in a wire is 0.72A. Calculate the charge that passes through the wire in; a. 4 s b. 60 s c. 180 s d. 7 s e. 0.5 s

Physics
1 answer:
Vlad [161]3 years ago
6 0
Let current be I, charge be Q and time be t.
Here we are provided with,
I = 0.72A
t = 4s / 60s / 180s / 7s / 0.5s
We know,
I = Q/t

Case I
---------
When, t = 4s
0.72 = Q/4
Q = 0.72 * 4 = 2.88C

Case II
----------
When, t = 60s
0.72 = Q/60
Q = 0.72 * 60 = 43.2C

Case III
-----------
When, t = 180s
0.72 = Q/180
Q = 0.72 * 180 = 129.6C

Case IV
-----------
When, t = 7s
0.72 = Q/7
Q = 0.72 * 7 = 5.04C

Case V
----------
When, t = 0.5s
0.72 = Q/0.5
Q = 0.72 * 0.5 = 0.36C
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Answer:

C

Explanation:

Sound waves speed up noticeably when moving through a solid or liquid, because all it is is just particles colliding; and particles are way closer together with those states of matter.

The speed of light can change when moving through different substances, but this is dependent on complicated factors such as frequency, polarization, intensity, et. cetera

The important part is that it does change speed, so your answer is C.

Hope this helps!

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3 years ago
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5 0
3 years ago
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The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the
8090 [49]
Compute first for the vertical motion, the formula is:

y = gt²/2 

0.810 m = (9.81 m/s²)(t)²/2 

t = 0.4064 s 


whereas the horizontal motion is computed by: 

x = (vx)t 

4.65 m = (vx)(0.4064 s) 

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed. 

(vy) = gt 

(vy) = (9.81 m/s²)(0.4064 s) 

(vy) = 3.99 m/s 


speed with which it hit the ground: 

v = sqrt[(vx)² + (vy)²] 

v = sqrt[(11.44 m/s)² + (3.99 m/s)²] 

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6 0
3 years ago
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

  • f is the focal length of the mirror
  • v is the object distance
  • u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

#SPJ1

7 0
2 years ago
2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n
AleksandrR [38]

Answer:

No

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}

For the ships to sight each other, distance must be 5 or smaller

d \leq 5

\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5

(12 - 12t)^2 + (9t)^2 \leq 25

144t^2 - 288t + 144 + 81t^2 - 25 \leq 0

225t^2 - 288t + 119 \leq 0

(15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0

(15t^2 - 9.6)^2 + 26.84 \leq 0

Since (15t^2 - 9.6)^2 \geq 0 then

(15t^2 - 9.6)^2 + 26.84 > 0

So our equation has no solution, the answer is no, the 2 ships never sight each other.

8 0
3 years ago
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