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Anna [14]
3 years ago
6

What is the temperature outside of a tree?

Physics
1 answer:
ozzi3 years ago
7 0
About 21c because it also depends on the weather outside
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Tasya [4]

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The poles of the magnetic field generated around the armature are attracted to the opposite poles of the permanent magnet. As the opposite poles align, the commutator reverses the current direction so like poles are aligned and the armature continues to spin.

Explanation:

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The lithosphere contains rocks, soils, and minerals.
sergij07 [2.7K]

Answer: TRUE

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A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a
Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

x_t(t)=d+\frac{1}{2}a_tt^2

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

a_t=2.10 m/s^2 is the acceleration of the truck

The car position instead it is given by the equation

x_c(t)=\frac{1}{2}a_ct^2

where

a_c=3.40 m/s^2 is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

x_t(t') = d + 60

Therefore, solving the equation, we find the time t when  this occurs:

d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

x_c(t')=x_t(t')

And by solving the equation, we find the value of d, the initial distance between car and truck:

\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m

C)

In order to find the speed of each vehicle, we use the following suvat equation:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

a_t = 2.10 m/s^2

So its speed after t = 7.56 s is

v_t = 0+(2.10)(7.56)=15.9 m/s

For the car, we have

u = 0

a_c=3.40 m/s^2

So its speed after t = 7.56 s is

v_c=0+(3.40)(7.56)=25.7 m/s

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
4 years ago
A 5kg block of iron is heated to 800degree C.it is placed in a tub containing 2litre of water at 15degree C.Assuming all the wat
icang [17]
Te correct answer is a
6 0
3 years ago
It is night. Someone who is 4 feet tall is walking away from a street light at a rate of 8 feet per second. The street light is
KiRa [710]

Answer: 4 ft/s

Explanation:

Given

height of man=4 ft

speed of person v=8 ft/s

height if street light=12 ft

Let x be the distance between person and street light and y be the length of his shadow

From diagram

as the two triangle ADE and ABC are similar therefore we can say that

\frac{4}{12}=\frac{y}{x+y}

\frac{1}{3}=\frac{y}{x+y}

x+y=3y

x=2y

differentiate above Equation w.r.t time we get

\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{8}{2}=4 ft/s

6 0
3 years ago
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