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3 years ago

15

John bounces a ball off the ground, and it leaves the ground with a velocity of 10 m/s. How long will it take the ball to reach

maximum height?

2 answers:

Maksim231197 [3]3 years ago

8 0

Explanation:

This is true because at maximum height, the velocity is 0

wariber [46]3 years ago

3 0

Gravity makes an upward-moving object move 9.8 m/s slower every second.

So if it leaves the ground at 10 m/s, it'll run out of upward speed in

(10 m/s) / (9.8 m/s²) = <em>1.02 seconds </em>

At that time, since it has no more upward speed, it'll start falling. So at that instant, it was exactly at its maximum height.

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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

Soil conservation involves protecting _______ and preventing _______.

Naddika [18.5K]

<span>Soil conservation involves protecting soil quality and preventing erosion </span>

7 0

3 years ago

Read 2 more answers

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversion is Juan most likely doing

Semmy [17]

If Juan used a Celsius thermometer, it would tell him the Celsius temperature.

If he added 273 to that number, he'd have the "absolute" or Kelvin temperature.

7 0

2 years ago

Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec

STatiana [176]

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

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

Using the formula

Acceleration= $a=\frac{0-4.5}{0.5}$

deceleration=a= $-9m/s^2$

We know that

Force =ma

Using the formula and suppose mass of my body=m=40 kg

The force exerted by wall on you

Force= $40\times (9)=360N$

3 0

2 years ago