Ask question

Ask question

All categories

Lostsunrise [7]

3 years ago

11

Now Abel and Kato use what they learned to answer the following problem. The initial speed of a tennis ball is 54 m/s and the la

unch angle is θi = 21°. Neglect air resistance. What is the maximum height, h, of the tennis ball? m What is the range, R, of the tennis ball?

1 answer:

34kurt3 years ago

6 0

Answer:

h=19.4m

R=199.07 m

Explanation:

To solve this problem we use the parabolic motion equations:

We define:

$v_{i}$ total initial speed =54 $\frac{m}{s}$

$\alpha$ =angle that forms the total initial speed with the horizontal line= 21°

$v_{ix}$ :initial speed component in horizontal direction

= $v_{i} cos\alpha$ =54*cos 21= 50.41 m/s

$v_{iy}$ :initial speed component in vertical direction

= $v_{i} sin\alpha$ =54*sin21=19.35 m/s

$v_{x}$ :horizontal speed at any point on the parabolic path

$v_{y}$ : vertical speed at any point on the parabolic path

g= acceleration of gravity= 9,8 $\frac{m}{s^{2} }$

Equation of the speed of the football in the vertical direction :

$(v_{y} )^{2} =(v_{yi} )^{2} -2*g*y$ Equation (1)

Calculation of the maximum height(h)

The speed of the ball (vy) in the vertical direction gradually decreases until its value is zero when it reaches the maximum height.

We replace y=h, $v_{y} =0$ , $v_{iy} = 19.39\frac{m}{s}$ , $g=9.8\frac{m}{s^{2} }$ in the equation(1)

$0=19.35^{2} -2*9.8*h$

$h=\frac{19.35^{2} }{2*9.8}$

h=19.1 m

Calculating of the range (R)

Fórmula: $R=v_{ix} *t (m)$ Equation (2)

R is the maximum horizontal distance the ball reaches.

The time (t) for the ball to reach R is twice the time the ball spends to reach the maximum height. Then, we calculate the time $(t_{h} )$ ) when the ball reaches the maximum height

We apply the following equation to calculate $t_{h}$ :

$v_{y} =v_{iy} -g*t_{h}$

$0=19.35-9.8*t_{h}$

$9.8*t_{h} =19.35$

$t_{h} =\frac{19.35}{9.8}$

$t_{h} =1.97 s$

$t=2*t_{h}$

$t=3.95 s$

We replace $v_{ix} =50.41 \frac{m}{s}$ and $t=3.95\frac{m}{s}$ in the equation (2)

R=50.41*3.95

R=199.07 m

You might be interested in

How do you know the earth is rotating on its axis

cupoosta [38]

Answer:

Read below!

Explanation:

You can watch the sun wheel across the sky during the day, and the stars at night. Focus a telescope on any star besides the north star--especially southern stars--and you can watch them drift across your field of view.

An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.

As we know the stars are massive bodies of significant and varying distance to the earth, the notion they all swing around us in unison seems highly implausible

4 0

3 years ago

1. How much energy is needed just to melt 0.56kg of ice at 0◦ C?

ivann1987 [24]

Answer:

This will require 266.9 of heat energy.

Explanation:

To calculate the energy required to raise the temperature of any given substance, here's what you require:The mass of the material, m The temperature change that occurs, ΔT The specific heat capacity of the material,

c

(which you can look up). This is the amount of heat required to raise 1 gram of that substance by 1°C.

Here is a source of values of

c for different substances:

Once you have all that, this is the equation:

Q=m×c×ΔT(Q is usually used to symbolize that heat required in a case like this.)For water, the value of c is 4.186g°C So, Q=750×4.186×85=266=858=266.858

5 0

3 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

<em>where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

<em>where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass. </em>

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago

The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is

astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

$E = \frac{T}{P*sin(\theta)}= \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C$

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

6 0

3 years ago

Astronomers have found water on mars.

vodomira [7]

The thin atmosphere of Mars is thought to be due to the planet's lack of a magnetic field, which has allowed the Solar wind to blow away much of the gas the planet once had. Venus, despite still having a thick atmosphere of CO2, surprisingly has a similar problem

4 0

2 years ago