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Delicious77 [7]
2 years ago
10

A train travels a distance of 1,2 km between two stations with an average velocity of 43.2 km/h. During it's motion, at the time

t1=40s it moved accelerated, then at time t2 it moved uniformly, then at t3=40s it moved uniformly slowed. Find the maximum velocity of the train. The acceleration during the slowed motion is equal in modulus to the acceleration during the accelerated motion.
Physics
1 answer:
notsponge [240]2 years ago
7 0
For the answer to the question above,
The whole trip covered d = 1.2 km at 43.2 km/hr taking a total time of T = (1.2 / 43.2) hr = 100 s. Since T = t1 + t2 + t3 and t1=t3=40 s, then t2 = 100-40-40 = 20 s. 

<span>The area under the v vs. t curve is the distance d traveled. The shape of the curve is an isosceles trapezoid (trapezium in UK English) with bases T and t2 and height V. So, from the area formula for a trapezoid: </span>

<span>d = (V/2)(T + t2) </span>
<span>V = 2D / (T + t2) = (2.4 km) / (120 s) = 0.02 km/s </span>

<span>Multiply by 3600 s/hr to get 72 km/h to match the original units in the problem</span>
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weqwewe [10]
Arginine is a basic aminoacid, because it has two amino groups and one acid group. At a low pH, every ionizable group is protoned. At a little higher pH, the acid group looses its proton. A little higher pH, one amino group looses its proton. At a very high pH, all ionizable groups are not protoned.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> So 9.20 is higher tan the second pKa and lower than the third pka. This means the acid has already lost its proton, and one of the aminos too, but the second amino hasn’t. When an acid is not protoned, it has a negative charge. When an amino is not protoned, it’s neutral. When an amino is protoned, it has a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative charge (the acid), what makes it neutral.
4 0
2 years ago
A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N What is the value of t
nadya68 [22]

1) weight of the box: 980 N

The weight of the box is given by:

W=mg

where m=100.0 kg is the mass of the box, and g=9.8 m/s^2 is the acceleration due to gravity. Substituting in the formula, we find

W=(100.0 kg)(9.8 m/s^2)=980 N


2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

N=W cos \theta

where W is the weight of the box, calculated in the previous step, and \theta=50^{\circ} is the angle of the ramp. Substituting, we find

N=(980 N)(cos 50^{\circ})=630 N


3) Acceleration: 7.5 m/s^2

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

a_p = g sin \theta

Substituting, we find

W_p = (9.8 m/s^2)(sin 50^{\circ})=7.5 m/s^2

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2 years ago
A worker pushes a 7 kg shipping box along a roller track. Assume friction is small enough to be ignored because of the rollers.
Bingel [31]

Answer:

a) Fₓ = 23.5 N

b) Net force = Fₓ

Explanation:

An image of the question as described is attached to this solution.

From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)

a) From the diagram, the horizontal component of the force is

Fₓ = 25 cos 20° = 23.49 N = 25 N

b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain

Net force = Fₓ - Frictional force

But frictional force is 0 N

Net force = Fₓ

Hope this Helps!!!

6 0
2 years ago
A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul
Rashid [163]
The Young modulus is given by:
E= \frac{F /A}{\Delta L / L_0}
where
F is the force applied
L_0 is the initial length of the wire
A is the cross-sectional area of the wire
\Delta L is the stretch of the wire

The wire in the problem stretches by 0.5% of its length, this means 
\frac{\Delta L}{L_0}  = 0.005

We can also calculate the area of the wire; its radius is in fact half the diameter:
r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m
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A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

E=  \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2
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1 year ago
A negative oil droplet is held motionless in a millikan oil drop experiment. What happens if the switch is opened?
scoray [572]

In Millikan oil drop experiment, when the switch is opened and by altering supply the charge of electron is determined.

Explanation:

Millikan's oil drop experiment is held to determine the terminal velocity and charge of the oil drop.

Firstly without any supply of voltage when an oil drop is sprinkled and these droplets gather electrons together and gives negative charge as they pass through air.

By applying and altering voltage applied on the plates, drop can be suspended in air. Millikan observed one drop after another, varying the voltage and noting the effect. After many repetitions he concluded that charge could assume only certain fixed values.

After conducting many times he concluded 1.602176487 ×10−19 C as the charge of an electron.

0 0
1 year ago
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