When voltage increases and resistance remains constant what must happen to current?

A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the b

Mariulka [41]

We have centripetal acceleration = $v^2/r = 2g$

So, v = $\sqrt{2gr}$

Now by equation of motion we have S= ut +0.5a $t^2$

S =displacement, u = initial velocity, a= acceleration and t = time

Here S = 2r and a = g , u = 0

$2r = 0+\frac{1}{2} *g*t^2$

$t = \sqrt{\frac{4r}{g} }$

Distance traveled in horizontal direction = $\sqrt{2gr}*\sqrt{\frac{4r}{g} }= \sqrt{8r^2} =2r\sqrt{2}$

8 0

4 years ago

One billiard ball is shot east at 2.2m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing co

Kruka [31]

Answer:

$v_{1}=1.886 \frac{m}{s}$

β= 57.99 south of east

Explanation:

$v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=$ ?

Velocity in axis x the two balls come one from east and west

$m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\ v_{fx1}=1 \frac{m}{s} \\$

Velocity in axis y initial is zero so:

$v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}$

$v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}$

Angle is find using:

tan(β)= $\frac{v_{fy}}{v_{fx}}$

$\beta =tan^{-1}*\frac{1.6}{1}=57.99$

5 0

4 years ago

Someone please help asap

Minchanka [31]

Huhhhh this is confusing ummm hello I guess

7 0

3 years ago

Click to review the online content. Then answer the question(s) below, using complete sentences. Scroll down to view additional

Serga [27]

Answer:There is growing interest and evidence that sleep loss and sleep disorders have a significant impact on metabolism. Laboratory studies have clearly shown that sleep deprivation can alter the glucose metabolism and hormones involved in regulating metabolism, that is, decreased leptin levels and increased ghrelin levels.Aug 2, 2010

Explanation:

4 0

3 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

4 years ago