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A ball is launched from ground level at 20 m/s at an angle of 40° above the

DedPeter [7]

(a) The ball's height y at time t is given by

y = (20 m/s) sin(40º) t - 1/2 g t ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :

0 = (20 m/s) sin(40º) t - 1/2 g t ²

0 = t ((20 m/s) sin(40º) - 1/2 g t )

t = 0 or (20 m/s) sin(40º) - 1/2 g t = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 g t

t = (40 m/s) sin(40º) / g

t ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So

0² - ((20 m/s) sin(40º))² = 2 (-g) y

where y in this equation refers to the maximum height of the ball. Solve for y :

y = ((20 m/s) sin(40º))² / (2g)

y ≈ 8.4 m

8 0

3 years ago

compare your previous result to the present hrf result are there any changes if yes explain your answer

garri49 [273]

Answer:

I don't know what are you saying but I don't have any results

Explanation:

$kai6417$

4 0

2 years ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0

3 years ago

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0

3 years ago

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

3 years ago