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Alona [7]
3 years ago
12

A train travels 90 kilometers in 4 hours, and then 82 kilometers in 1 hours. What is its average speed?

Physics
1 answer:
kvasek [131]3 years ago
5 0
Average speed = total distance / total time.

That's 172 km / 5 hr = 34.4 km//hr
You might be interested in
Which object would be the best insulator of electricity?
solniwko [45]

Answer:

a rubber band

Explanation:

hope this helps!

7 0
2 years ago
Read 2 more answers
A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc
katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

cos \theta = \frac {V_{river}}{V_{boat}}

When solving for theta, we get that:

\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})

so now we can substitute the corresponding values:

\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})

Which yields:

\theta = 60^{o}

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

V_{y}=5.60km/hr*cos(210^{o})

V_{y}=-4.85km/hr

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

t=\frac{y}{V_{y}}

t=\frac{5.60km}{4.85km/hr}=1.155hr

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

V_{ds}=V_{river}+V{boat}

V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr

and now up stream:

V_{us}=V_{boat}-V{river}

V_{us}=5.60km/hr-2.80km/hr=2.80km/hr

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

t_{ds}=\frac{x}{v_{ds}}

t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr

and the time up stream:

t_{us}=\frac{x}{v_{us}}

t_{us}=\frac{2.80km}{2,80km/hr}=1hr

so the total time will be:

t_{ds}+t_{us}=0.333hr+1hr=1.333hr

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

t=\frac{d}{v_{boat}}

t=\frac{5.60km}{5.60km/hr}

so

t= 1hr

4 0
3 years ago
Read 2 more answers
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
If we decrease the amount of force, and kelp all other factors the same, what will happen to the amount of work?
Artist 52 [7]

Answer:

Distance will decrease and work will decrease:

F = m a      Newton's Second Law

a = F / m       decreasing force will decrease acceleration

S = 1/2 a t^2 = 1/2 (F / m) t^2     distance traveled will decrease as force        decreases

W = F * S       work will decrease as both force and distance decrease

8 0
3 years ago
A traffic light of weight 100 N is supported by two ropes that make 30-degree angle with the horizontal. What is the vertical co
alina1380 [7]

Answer:

T_{vertical} = 50 N

Explanation:

given,

traffic light weight = 100 N

angle at which the rope is supported = 30°

vertical component of force = ?

2 T sin \theta= W

2 T sin 30^0= 100

T = \dfrac{100}{2 sin 30^0}

T = 100\ N

T_{vertical} = T sin 30^0

T_{vertical} = 100\times \dfrac{1}{2}

T_{vertical} = 50 N

5 0
3 years ago
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