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Zigmanuir [339]
3 years ago
14

What are some examples where the SI is used everyday in the United States.

Physics
1 answer:
schepotkina [342]3 years ago
8 0

In any scientific use.

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A circuit is built based on the circuit diagram shown. What is the current in the 50 Ω resistor
In-s [12.5K]

Answer:

1.2 A

Explanation:

From the diagram attached, The three resistors are parallel because the each ends of the resistors are connected together. Since they are in parallel, the voltage across each resistor is the same. The voltage source connected in parallel to the resistors is 60 V. Therefore the voltage across the 50 Ω resistor is 60 V. Using ohm law:

Voltage (V) = Current (I) × Resistance (R)

V = IR

I = V/R

I = 60 V/ 50 Ω

I = 1.2 A

The current in the 50 Ω resistor is 1.2 A

5 0
3 years ago
9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?
kakasveta [241]

Answer:

Option B (1.51 m)

Explanation:

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5 0
2 years ago
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Question 3 Please, i need help! Thank you
Tom [10]
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4 0
3 years ago
A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball
DerKrebs [107]
A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N

.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N

5 0
3 years ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
3 years ago
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