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lisabon 2012 [21]

4 years ago

8

A race car accelerates on a straight road from rest to speed of 180 km/hr in 25s . Assuming uniform acceleration of the car thro

ugh out find the find the distance covered in this time .

1 answer:

adelina 88 [10]4 years ago

3 0

X-x0=0.5(v0+v)t ergo x=0.5(50)25=625m

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9) An electrical appliance has a resistance of 25 N. When this electrical ap-

Neko [114]

(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

= 9.2 A

6 0

3 years ago

Increasing the speed of an object blank its potential energy

ella [17]

There is no direct connection between the speed of an object and its potential energy.

UNLESS the object is falling, and it's GETTING its speed FROM its potential energy. If that's what's going on, then of course if the speed increases, the object's kinetic energy increases, and that increase has to come from using up some of the potential energy.

7 0

3 years ago

V125BC [204]

Answer:

her arm when she is throwing it

Explanation:

because that is the action

7 0

3 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

4 years ago

If we're interested in knowing the rate at which light energy is received by a unit of area on a particular surface, we're reall

frosja888 [35]

Answer: C. illuminance of a surface.

a) Incandescence: The phenomenon of light emission by a body as a result of high temperature.

B. Luminous flux : It is the quantity of the energy of the light emitted per second in all directions.

C. Illuminance of a surface :describes the quantity of light emitted by a light source or received at a surface.

D. luminous intensity : the quantity of visible light that a point source radiates in a given direction.

4 0

4 years ago