All categories

2 years ago

Question 11 of 29

Physics

1 answer:

lara [203]2 years ago

3 0

Answer:

the answer is a

Explanation:

wawawawawawawawawawawawawa

You might be interested in

Select whether the argument is an example of a deductive or an inductive argument:

viva [34]

The answer is a inductive

7 0

3 years ago

An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.

horrorfan [7]

Answer:

f = 12 cm

Explanation:

Center of Curvature:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.

The Radius of Curvature:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.

Focal Length:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

$f = \frac{R}{2}$

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

$f = \frac{24\ cm}{2}$

f = 12 cm

8 0

2 years ago

A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc

Hoochie [10]

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v= 0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

$a=- \dfrac{20^2}{2\times 30} \ m/s^2$

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be - 6.667 m/s² .

6 0

2 years ago

Please help me with this physics prooblem

zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

$x=v_0\cos20.0^\circ t+\dfrac12a_xt^2$

$y=v_0\sin20.0^\circ t+\dfrac12a_yt^2$

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

$x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ$

$y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ$

So we have enough information to solve for the components of the acceleration vector, $a_x$ and $a_y$ :

$x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}$

$y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}$