What is the relationship between the force and distance on a balances lever?

What is the relationship between resistance and current.

Tasya [4]

Answer:

They are inversely proportional with each other. $R$ ∝ $\frac{1}{I}$

Explanation:

By the equation V=IR, assume V is constant.

Since I and R are on the same side,

if I increase, R decreases, or the vice versa.

7 0

3 years ago

A 2,100 kg car is lifted by a pulley. If the cable breaks at 4.50 m, what is the velocity of the car when it hits the ground

kirill [66]

The problem involves the conversion of potential energy to kinetic energy as the object falls from rest. Energy is conserved, so the equation used is:

PEi + KEi = PEf + KEf

Since the object is falling from rest, the initial kinetic energy is zero. Also, since the object hits the ground at its final position, the final potential energy is zero. This leaves:

PEi = KEf
mgh = 1/2 mv^2

*cancel out mass on both sides of the equation

gh = 0.5v^2
v = sqrt(2gh) = sqrt(2*9.81*4.5) = 9.40 m/s --> final ans.

4 0

4 years ago

QUESTION 2 DOK 3 A Thompson's gazelle has a maximum acceleration of 4.5 m/s2 At this acceleration, how much time is required for

Dominik [7]

Answer:

2.47 s

Explanation:

Convert the final velocity to m/s.

40 km/h → 11.1111 m/s

We have the acceleration of the gazelle, 4.5 m/s².

We can assume the gazelle starts at an initial velocity of 0 m/s in order to determine how much time it requires to reach a final velocity of 11.1111 m/s.

We want to find the time t.

Find the constant acceleration equation that contains all four of these variables.

v = v₀ + at

Substitute the known values into the equation.

11.1111 = 0 + (4.5)t
11.1111 = 4.5t
t = 2.469133333

The Thompson's gazelle requires a time of 2.47 s to reach a speed of 40 km/h (11.1111 m/s).

5 0

3 years ago

A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused

Nonamiya [84]

Answer:

The velocity $\vec{v}_{c/g}$ of the cart with respect to the ground is

$\vec{v}_{c/g}=-40\hat{x}+80\hat{y}\, km/h$

if we consider North the positive y-direction and East the positive x-direction.

Explanation:

We have for relative motion the following expression:

$\vec{v}_{c/g}=\vec{v}_{c/p}+\vec{v}_{p/g}$

Where $\vec{v}_{c/g}$ is the velocity of the cart with respect to the ground, $\vec{v}_{c/p}$ is the velocity of the cart with respect to the plane and $\vec{v}_{p/g}$ is the velocity of the plane with respect to the ground.

We find that:

$\vec{v}_{c/p}=-20\hat{y}$

$\vec{v}_{p/g}=-40\hat{x}+100\hat{y}$

Thus:

$\vec{v}_{c/g}=-20\hat{y}-40\hat{x}+100\hat{y}=-40\hat{x}+80\hat{y} \, km/h$

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:

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