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3 years ago

A ball is kicked from the top of a building with a velocity of 50 m/s and lands 165 m away from the base of the buildi

Physics

1 answer:

solniwko [45]3 years ago

5 0

Answer:

32.3 m/s

Explanation:

The ball follows a projectile motion, where:

- The horizontal motion is a uniform motion at costant speed

- The vertical motion is a free fall motion (constant acceleration)

We start by analyzing the horizontal motion. The ball travels horizontally at constant speed of

$v_x = 50 m/s$

and it covers a distance of

d = 165 m

So, the total time of flight of the ball is

$t=\frac{d}{v_x}=\frac{165}{50}=3.3 s$

In order to find the vertical velocity of the ball, we have now to analyze its vertical motion.

The vertical motion is a free-fall motion, so the ball is falling at constant acceleration; therefore we can use the following suvat equation:

$v_y = u_y +at$

where

$v_y$ is the vertical velocity at time t

$u_y=0$ is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity (taking downward as positive direction)

Substituting t = 3.3 s (the time of flight), we find the final vertical velocity of the ball:

$v=0 + (9.8)(3.3)=32.3 m/s$

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Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

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3 years ago

If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

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Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

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Now according to the question the distance is increased by three times than,

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Therefore,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }$

Therefore the intensity will become 1/9 times to the initial intensity.

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3 years ago

Question I. SUHU ULT

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Explanation:

Feathers are great thermal insulators. The loose structure of down feathers traps air.

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3 years ago

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Answer:

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3 years ago

A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh

irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

$K=\frac{1}{2}mv^2$

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

$U=mgh$

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

$a=\frac{v-u}{t}$

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

$u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s$

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

$\Delta v = 0 - (+9.8 m/s)=-9.8 m/s$

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

$a=\frac{v-u}{t}$

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

$v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s$

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

$\Delta v = v -u$

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - 0=-9.8 m/s$

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

$\Delta v = v -u$

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

$\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s$

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

$mg = ma$

which means that the acceleration is

$a= g = -9.8 m/s^2$

and the negative sign means it points downward.

7 0

3 years ago