Ask question

Ask question

All categories

2 years ago

7

Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t

he grass causes the ball to decelerate at a constant rate of 1.0 m/s2, how long does it take for the ball to reach you?

1 answer:

LekaFEV [45]2 years ago

3 0

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

Initial velocity = 16 m/s

Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

$x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}$

Substituting into the equation, we have;

$x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}$

$x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}$

$x = \frac {-32 \pm \sqrt {1024 + 800}}{2}$

$x = \frac {-32 \pm \sqrt {1824}}{2}$

$x = \frac {-32 \pm 42.71}{2}$

$x_{1} = \frac {-32 + 42.71}{2}$

$x_{1} = \frac {10.71}{2}$

x1 = 5.355

We do not need the negative value of x, so we proceed.

Therefore, time = 5.355 seconds

You might be interested in

1.is anyone else bored with school<br><br> 2. is anyone getting out of school soon or at 3

storchak [24]

Answer:

i am bored and i get out in 10 minuets

Explanation:

3 0

2 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho

Evgesh-ka [11]

Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

b) $y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

For this case we have the following data:

x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

Part b

For this case we have the following trend equation given:

$y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

7 0

3 years ago

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f

tatiyna

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_{f}$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_{cm }^{2}$ + ½ $I_{cm}$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

4 0

3 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago