Which has a larger resistance a 60 w lightbulb or a 100 w lightbulb?

Which of the following diagrams involves a virtual image ?

sergiy2304 [10]

Answer:

The third diagram

Explanation:

A virtual image is an image that can not be formed on a screen.
A convex lens can form both virtual and real image depending on the position of the object from the lens.
A virtual image in convex lens is formed when the object is placed between the focus and the optical center of the lens.
In the third diagram, a virtual image is formed because the position of the object is between the focus and the optical center of the convex lens.

7 0

2 years ago

A graph titled Distance as a Function of Time with horizontal axis time (seconds) and vertical axis distance (meters). A straigh

yanalaym [24]

Answer:

3 m

Explanation:

6 0

1 year ago

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A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His

adoni [48]

Answer:

1) $v_2=23\ m.s^{-1}$ & $x_2=43\ m$ east of sign post

2) $x'=55\ m$ east of sign post

3) $x_n=205\ m$ east of the signpost.

4) $v_z=35\ m.s^{-1}$

Explanation:

Given:

position of motorcyclist on entering the city at the signpost, $x_0=0\ m$

time of observation after being at x=5m east of the signpost, $t_m=0\ s$

constant acceleration of the on entering the city, $a=4\ m.s^{-2}$

distance of the motorcyclist moments later after entering, $s_m=5\ m$

velocity of the motorcyclist moments later after entering, $u_m=15\ m.s^{-1}$

Now the initial velocity on at the sign board:

$u_m^2=u^2+2.a.x_m$

where:

$u=$ initial velocity of entering the city at the signpost

Putting respective values:

$15^2=u^2+2\times 4\times 5$

$u=13.6015\ m.s^{-1}$

1)

Position at time $t_2=2\ s$ sec.:

Using equation of motion,

$x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5$ because it has already covered 5m before that point

$x_2=15\times 2+0.5\times 4\times 2^2+5$

$x_2=43\ m$ east of sign post

Velocity at time $t_2=2\ s$ sec.:

$v_2=u_m+a.t_2$

$v_2=15+4\times 2$

$v_2=23\ m.s^{-1}$

2)

Position when the velocity is $v'=25\ m.s^{-1}$ :

using equation of motion,

$v'^2=u_m^2+2.a.x'+5$

$25^2=15^2+2\times 4\times x'+5$

$x'=55\ m$ east of sign post

3)

Given that:

acceleration be, $a_n=2\ m.s^{-2}$

time, $t_n=5\ s$

Position after the new acceleration and the new given time:

using equation of motion,

$x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5$

$x_n=15\times 5+0.5\times 2\times 5^2+5$

$x_n=205\ m$ east of the signpost.

4)

now time of observation, $t_z=5\ s$

$v_z=u_m+a.t_z$

$v_z=15+4\times 5$

$v_z=35\ m.s^{-1}$

8 0

2 years ago

Zamir and Talia raced through a maze. Zamir walked 2 m north, 2 m east, 4 m south, 2 m east, 4 m north, 2 m east, 3 m south, 4 m

Alja [10]

Answer : Zamir's displacement and Talia's displacement is equal.

Explanation :

Displacement is explained to be the changing position of an object.

Zamir covers total distance 27 m and Talia covers total distance 19 m but Zamir's initial and final position and Talia's initial and final position is same.

So, we can say that Zamir's displacement and Talia's displacement is equal.

6 0

2 years ago

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On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car c

Nana76 [90]

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

$F=\mu mg = m \frac{v^2}{r}$ (1)