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Darya [45]
2 years ago
8

1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00

Physics
1 answer:
rosijanka [135]2 years ago
8 0

12.00 min = 0.2 hr

8.00 min = 0.15 hr

Total distance:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)

= 8.25 km

Average speed:

(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3

= 15 km/hr

Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

= 0.25 km

Average velocity:

(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3

≈ 1.67 m/s

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An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.
bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

                        p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

                      θ = 0º

                      U = 1.16 10⁻¹⁴ 1160 cos 0

                      U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

                       θ = 180º

                      cos 180 = -1

                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

6 0
3 years ago
A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
An over-caffeinated student stands on a table which weighs 600 newtons. The student has a mass of 50 kg. What is the weight of t
liberstina [14]

Answer:50kg

Explanation:

6 0
3 years ago
Ai là người phát hiện trái đất hình cầu đầu tiên ?
Scorpion4ik [409]

Answer:

Can't understand the language

5 0
2 years ago
How many pounds of force must a guardrail’s top rail be able to withstand? A. 150 lbs. B. 200 lbs. C. 400 lbs. D. 5,000 lbs.
Greeley [361]
At least 200 pounds.
hope this one helps
3 0
3 years ago
Read 2 more answers
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