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2 years ago

A ball is traveling 24° above the horizontal at a speed of 12 m/s. What is the vertical component of its speed?

Physics

1 answer:

victus00 [196]2 years ago

3 0

Answer:

4.88 m/s

Explanation:

Vertical component would be 12 * sin 24 = 4.88 m/s

Horizontal is 12 * cos 24

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A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box t

Marrrta [24]

Answer:v=3.08 m/s

Explanation:

Given

mass of student $m=21 kg$

distance moved $d=10 m$

Force applied $F=10 N$

acceleration of system during application of force is a

$a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2$

using $v^2-u^2=2 as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-0=2\times 0.476\times 10$

$v=\sqrt{9.52}$

$v=3.08 m/s$

6 0

3 years ago

8. The legs of a young man are each 0.650 meters long. What is his maximum walking speed?

Norma-Jean [14]

Answer:

2.52 m/s

Explanation:

When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.

Sum of the forces in the radial direction:

∑F = ma

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 0.650 m:

v = √(9.8 m/s² × 0.650 m)

v = 2.52 m/s

8 0

3 years ago

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

3 years ago

WARNING IF YOU CANT ANSWER IT DONT EVEN ANSWER OR IM REPORTING AND POSTING YOUR ANSWER ONLINE!!!!! (⌐■_■)☜(⌒▽⌒)☞

andrezito [222]

Answer:

the answer is false :)

Explanation:

Science is based on theories and tests, if someone has a theory they will test and if it is correct yippie! and if it isn't they will redo the test over until they get it correct or unless it is unsolvable.

3 0

3 years ago

Read 2 more answers

An object travels 20 m in 10s What is its speed?

ella [17]

speed=distance/time

distance=20m

time=10s

speed=?

speed=20×10

speed=200m/s

3 0

3 years ago