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Gekata [30.6K]
3 years ago
13

Please guys help me with this question.​ please show the working

Physics
1 answer:
yawa3891 [41]3 years ago
8 0

Answer:

lolllll

Explanation:

looobsbsbjwkskkskwk else's sjwbs w ksj

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a truck is moving around a circular curve @ a uniform velocity of 13m/s. if the centifrical force on truck is 3,300N with truck
VMariaS [17]

<u>Answer</u>

81.94 m


<u>Explanation</u>

The centripetal force of an object moving in a circular path is given by:

F = mv²/r  Where m is the mass of the object, v is the constant velocity and r is the radius of the curve.

F = mv²/r

3,300 = (1600×13²)/r

3,300 = 270,400/r

r = 270,400/3,300

  = 81.94 m

8 0
3 years ago
A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu
loris [4]
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k  · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²

We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m

Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
      = </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
      = 39.6 N/C

Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
4 0
4 years ago
What is space mostly composed of?
miv72 [106K]
Nothing it nor breatheable you could die and it cold
8 0
3 years ago
If a car drives 10 miles NE and then due East for 10 minutes traveling 8
Nimfa-mama [501]
I think the answer is 28 miles
7 0
2 years ago
Read 2 more answers
I need help with one through six please
dybincka [34]

Answer:

1] 8500000 = <u>8.5 × 10⁶</u>

2] .000072 = <u>7.2 × 10⁻⁵</u>

3] 5.3 × 10⁴ = <u>53000</u>

4] 2.8 × 10⁻³ = <u>0.0028</u>

5] Velocity = \frac{distance}{time}

 V = \frac{50}{10}

 <u>V = 5 m/s</u>

6] Acceleration = \frac{V1-V2}{time}

 A = \frac{30-15}{3}

 A = \frac{15}{3}

 <u>A = 3 m/s²</u>

7 0
2 years ago
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