All categories

3 years ago

Which of these is the most effective way for Leanna to cool down after an intense bike ride

Physics

1 answer:

Sonja [21]3 years ago

6 0

I am pretty sure the answer would be too stretch

You might be interested in

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

3 years ago

6) A boat crosses a river at a constant engine speed of 2.0 m/s under pointed Directly west. The river runs directly south at 2.

TEA [102]

Answer:ligma ;)

Explanation:

8 0

2 years ago

When you are high up in the air you<br> have<br> greater potential energy<br> less potential energy

mina [271]

Answer:

Its Greater potential energy because the air is high up and that makes high energy power

Explanation:

5 0

3 years ago

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:

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:

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

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:

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

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

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

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