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Ira Lisetskai [31]
3 years ago
12

Which measurement is a potential difference? A.115J B.115V C.115N D.115C

Physics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

B

Explanation:

Potential difference has a SI Unit of Volt and its symbol is <em>V</em>. Hence answer is <u>B</u>.

A is wrong as it has the unit Joule <em>(J)</em> which is the SI unit for energy.

C is wrong as it has the unit Newton <em>(N)</em> which is the SI unit for force.

D is wrong as it has the unit Coulomb <em>(C)</em> which is the SI unit of charge.

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When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is
lukranit [14]

Answer:

The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°

Explanation:

According to Snell's Law,

n₁ sin θ₁ = n₂ sin θ₂

When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n

n sin θ₁ = n sin (90° - θ₁)

Note that from trigonometric relations,

Sin (90° - θ₁) = cos θ₁

n sin θ₁ = n cos θ₁

(sin θ₁)/(cos θ₁) = 1

tan θ₁ = 1

θ₁ = arctan 1 = 45°

Hope this Helps!!!

7 0
3 years ago
Read 2 more answers
Pyramid math (algebra 1)
mixas84 [53]
5436 7699 2360 3488 2113
3 0
3 years ago
A diagram of a closed circuit with a power source on the left labeled 6 V. There are 3 resistors in parallel, separate paths, co
AfilCa [17]

Answer:

Current: 1.0 Amperes

The minimum current is flowing through path D

Explanation:

We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

\frac{1}{R_e} =\frac{1}{R_B}+\frac{1}{R_C}+\frac{1}{R_D}\\\frac{1}{R_e} =\frac{1}{10}+\frac{1}{20}+\frac{1}{50}=0.17\\R_e=(1/0.17)\Omega\\R_e=5.88 \Omega

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

V=I*R\\6V=I*5.88\Omega\\I=\frac{6}{5.88} Amp\\I=1.02A

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.

The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.

Than means that the lowest current will be registered through branch D where the 50 \Omega resistor is.

8 0
2 years ago
Read 2 more answers
Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m
ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

 a = 62.5²/(2 × 22)

 a = 88.78m/s²

the time you need to get this speed

     v = v₀ + a t

     t = v / a

     t = 62.5 / 88.78

     t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

     t = 1,025 s

      a = 55.43 m / s²

4 0
2 years ago
A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal. Just before it h
Nutka1998 [239]

Answer:

the change in thermal energy of the projectile is 43.8 kJ

Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final  velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)

KE =  ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ

8 0
3 years ago
Read 2 more answers
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