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Natalija [7]
3 years ago
7

Can someone PLEASE help me with this. Anyone that’s good in physics pls

Physics
1 answer:
kolbaska11 [484]3 years ago
4 0

Answer:

RT = 35 Ohms

Explanation:

Let the three resistors be R1, R2 and R3 respectively.

Given the following data;

R1 = 14 Ohms

R2 = 12 Ohms

R3 = 9 Ohms

To find the total resistance, RT;

Since all the resistors are connected in series, the total resistance is equal to the sum of all three resistance of the resistors.

RT = R1 + R2 + R3

Substituting into the equation, we have

RT = 14 + 12 + 9

RT = 35 Ohms.

Therefore, the total resistance of the circuit is 35 Ohms.

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A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
Teachers are interested in knowing what study techniques their students are utilizing. The researchers randomly select every 10t
uranmaximum [27]

Answer:

Simple Random Sample (SRS)

Explanation:

5 0
3 years ago
A student starts a food fight by throwing a 0.5 kg burrito at some girl he likes. He throws it kind-of hard so he accelerates it
Elenna [48]

Explanation:

m = mass of burrito thrown by the student = 0.5 kg

a = acceleration of the burrito thrown by the student = 3 m/s²

F = force applied by the student on the burrito = ?

According to newton's second law , the net force on an object is the product of its mass and acceleration. it is given as

F = ma

inserting the values

F = (0.5) (3)

F = 1.5 N

hence the net force on the burrito comes out to be 1.5 N

4 0
3 years ago
You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio
Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

E = \dfrac{\sigma }{2 \varepsilon_o}

E = \dfrac{8 \times 10^{-6} }{2 \times  8.85 \times 10^{-12}}

E = 4.51977 \times 10^5 \ V/m

On the square plate; The electric force F = Eq

F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)

F = 1.3559 \times 10^{-2} \ N

The acceleration a =\dfrac{ F}{m}{

a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}

a = 2.25988 \times 10^6 \ m/s^2

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

(\dfrac{1}{2}) mv^2 = \Delta Vq

v^2 = \dfrac{2 Eq}{dm}

v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }

v^2 = 64568142.86  \ m/s

v =\sqrt{ 64568142.86  \ m/s}

\mathbf{v = 8.035 \times 10^3 \ m/s}

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

6 0
3 years ago
If the speed of sound in air at 20 °C is 342m/s , what will be the increase in speed at 30 °C? with steps
andrew-mc [135]

Answer:

Hmm

Explanation:

I don't know sorry forgive me.

6 0
3 years ago
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