PHYSICS COLLEGE

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer 1

Answer:

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

$Speed = (Distance)/(Time)$

$Speed = (5)/(0.504)$

= 9.92 m/s

Now

$v^2 - u^2 = 2* g* h$

$9.92^2 = 2* 9.98 * h$

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Answer 2

Answer:

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

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A 50 kg child on a skateboard experiences a 75-N force as shown.What is the expected acceleration of the child? F = 75 N 7 . A. 0.67 m/s2 B. 1.50 m/s2 C. 6.70 m/s2 D. 25.0 m/s2
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"At time t = 0 a 2330-kg rocket in outer space fires an engine that exerts" an increasing force on it in the +x-direction. This force obeys the equation Fx=At2, where t is time, and has a magnitude of 781.25 N when t = 1.27 s .What impulse does the engine exert on the rocket during the 1.50- s interval starting 2.00 s after the engine is fired?

Answers

Answer:

$Imp = 5626.488\,(kg\cdot m)/(s)$

Explanation:

First, it is required to model the function that models the increasing force in the +x direction:

$a =(781..25\,N)/((1.27\,s)^(2))$

$a = 484 (N)/(s^(2))$

The equation is:

$F_(x) = 484\,(N)/(s^(2))\cdot t^(2)$

The impulse done by the engine is given by the following integral:

$Imp=484\,(N)/(s^(2)) \int\limits^(3.50\,s)_(2\,s) {t^(2)} \, dt$

$Imp = 161.333\,(N)/(s^(2))\cdot [(3.50\,s)^(3)-(2\,s)^(3)]$

$Imp = 5626.488\,(kg\cdot m)/(s)$

The mass of a baseball is 0.145 kg and its acceleration as it falls to the ground is 9.81 m/s2. How much force is acting on the baseball

Answers

Answer:

The answer is 1.42 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

Force = 0.145 × 9.81 = 1.42245

We have the final answer as

1.42 N

Hope this helps you

Which of these 23rd chromosomecombinations is likeliest to result in a
person with male and female traits?
ΧΟ
XXX
XXY
XY

Answers

Sorry if I’m wrong but I think it’s XO since o is not a sex chromosome

Express the following speeds as a function of the speed of light, c: (a) an automobile speed (93 km/h) (b) the speed of sound (329 m/s) (c) the escape velocity of a rocket from the Earth's surface (12.1 km/s) (d) the orbital speed of the Earth about the Sun (Sun-Earth distance 1.5×108 km).

Answers

Answer:

(a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

Explanation:

Given that,

Express the following speeds as a function of the speed of light.

Automobile speed = 93 km/h

We know that,

A function of speed of light is

$c=3*10^(8)\ m/s$

(a). Automobile speed = 93 km/h

Speed $v_(a)=93*(5)/(18)$

$v_(a)=25.83\ m/s$

We need to express the speed of automobile speed as a function of speed of light

Using formula of speed

$v=(v_(a))/(v_(l))$

Put the value into the formula

$v=(25.83)/(3*10^(8))$

$v=8.61*10^(-8)\ m/s$

(b). The speed of sound is 329 m/s.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(s))/(v_(l))$

Put the value into the formula

$v=(329)/(3*10^(8))$

$v=10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket from the Earth's surface is 12.1 m/s

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(e))/(v_(l))$

Put the value into the formula

$v=(12.1)/(3*10^(8))$

$v=4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun

Distance = 1.5\times10^{8}[/tex]

We know that,

The sun rays reached on the earth in 8 min 20 sec.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(d)/(t)$

Put the value into the formula

$v=(1.5*10^(8)*1000)/(500)$

$v=3*10^(8)\ m/s$

Hence, (a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric field. If the field strength is 713.0 N/C find the electric Tiux through the surface A) 560 Nm2/C B) 620 N·m2/C C) 160 n N.m2/C D) 280 N.m2/C

Answers

Answer:

electric flux is 280 Nm²/C

so correct option is D 280 Nm²/C

Explanation:

radius r = 0.50 m

angle = 30 degree

field strength = 713 N/C

to find out

the electric flux through the surface

solution

we find here electric flux by given formula that is

electric flux = field strength × area× cos∅ .......1

here area = πr² = π(0.50)²

put here all value in equation 1

electric flux = field strength × area× cos∅

electric flux = 713 × π(0.50)² × cos60

we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk

so we use cos60

electric flux = 280 Nm²/C

so correct option is D 280 Nm²/C

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the direction of travel. Explain what your answer means in terms of the object’s energy.

Answers

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^(\circ)$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$

$W=50* 9* \ cos(135)$

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

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