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Contact [7]
3 years ago
10

Calculate the current in a circuit when the voltage of the circuit is 5 V and the circuit has a total resistance of 5 Ω.

Physics
1 answer:
seropon [69]3 years ago
8 0

the answer is 1a as rearrange gives I = v divided by r

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A projectile lands at the same height from which it was launched. which initial velocity will result
Serhud [2]

The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin θ.

The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.

If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.

Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

Learn more here; brainly.com/question/12870645

5 0
3 years ago
An underwater scuba diver sees the Sun at an apparent angle of 43.0° above the horizontal. What is the actual elevation angle of
Inessa05 [86]

Answer:

The actual elevation angle is 12.87 degrees

Explanation:

In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that \theta_2=90-43=47.

Using Snell's Law we can write:

n_1\sin(\theta_1)=n_2\sin(\theta_2)

\implies \sin(\theta_1)=\frac{n_2}{n_1}\sin(\theta_2),

Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.

We thus have:

\sin(\theta_1)=n_2\sin(\theta_2)=1.333\sin(47)

\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13. Calling \alpha the actual angle of elevation, we get from the picture that \alpha=90-77.13=12.97

7 0
3 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
3 years ago
If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th
Rufina [12.5K]
Based on internet sources, <span>the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a). 
</span>
<span>Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. </span>
<span>Acceleration = 1.6 cramwells/s^2 </span>
<span>Radius = 150 cramwells </span>
<span>t = sqrt(150/1.6) = 9.68 s.

I hope this helps.</span>
8 0
3 years ago
_________ are exchanged or shared during the formation of a chemical bond.
Thepotemich [5.8K]
<span>A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds; or through the sharing of electrons as in covalent bonds.

</span>
6 0
3 years ago
Read 2 more answers
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