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MAVERICK [17]
1 year ago
9

A truck covered 2/7 of a journey at an average speed of 40

Physics
2 answers:
slavikrds [6]1 year ago
8 0

Answer:8h

Explanation:

lapo4ka [179]1 year ago
4 0

Answer: 8 km/h

Explanation:

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A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi
Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

δ=1.18cm

6 0
2 years ago
Read 2 more answers
A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen
Brums [2.3K]
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I  attached we can see that:
\frac{h_i}{h_0}=\frac{d_i}{d_0}
Two triangles that I marked are similar and from this we get:
\frac{h_i}{h_0}=\frac{d_i-f}{f}
The image and the object must have the same height so we get:
\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f
This tells how far the screen should be from the lens. 
The position of the screen on the optical bench is:
S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm

8 0
1 year ago
You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
5 0
2 years ago
Calculate the wavelength of a neutron that has a velocity of 200. cm/s. (the mass of a neutron = 1.675 × 10–27 kg and h = 6.63 ×
Licemer1 [7]
<span>To find the wavelength of a neutron can be calculated by using the formula Wavelength=h/m x v Where h is planck's constant m=mass of neutron v= velocity of the particle By substituting the given values Wavelength= 6.63 × 10–34 j s / 1.675 × 10–27 kg x 2 m/s^-1 Wavelength of a neutron=1.979 x 10^-7 m</span>
6 0
2 years ago
Read 2 more answers
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