All mutations are favorable and increase the ability for the

The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from

Allushta [10]

Answer:

y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

let's substitute

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

0 = v₀ sin θ - 9.8 t

Let's write our system of equations

45 = v₀ cos θ t

10 = v₀ sin θ t - ½ 9.8 t²

0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

v₀ sin θ = 9.8 t

we substitute

10 = (9.8 t) t - ½ 9.8 t2

10 = ½ 9.8 t2

10 = 4.9 t2

t = √ (10 / 4.9)

t = 1,429 s

Now let's use the first equation and the last one

45 = v₀ cos θ t

0 = v₀ sin θ - 9.8 t

9.8 t = v₀ sin θ

45 / t = v₀ cos θ

we divide

9.8t / (45 / t) = tan θ

tan θ = 9.8 t² / 45

θ = tan⁻¹ ( 9.8 t² / 45 )

θ = tan⁻¹ (0.4447)

θ = 24º

Now we can calculate the maximum height

v_y² = $v_{oy}^2$ - 2 g y

vy = 0

y = v_{oy}^2 / 2g

y = (20 sin 24)²/2 9.8

y = 3,376 m

the other angle that gives the same result is

θ‘= 90 - θ

θ' = 90 -24

θ'= 66'

for this angle the maximum height is

y = v_{oy}^2 / 2g

y = (20 sin 66)²/2 9.8

y = 17 m

thisis the correct

6 0

3 years ago

calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm

Ghella [55]

Answer:

R = 0.21 Ω

Explanation:

the formula:

R = r x l/A

R = (44 x 10-⁸ Ωm) x 1.5 / (π x (1 x 10-³ m)²)

R = 6.6 x 10-⁷ / 3.14 x 10-⁶

R = 0.21 Ω

8 0

3 years ago

You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

$A_2=\frac{5}{8}^2*\pi=1.227 in^2$

Re-arrange the equation to find the first ratio of rates we have:

$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

$\frac{V2}{V1}=0.640$

3 0

4 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

3 years ago

The density of sodium is 968 g/cm3. Suppose you need 250 gm of sodium for an experiment. Which volume of sodium do you need?

Masja [62]

The formula for the density of a substance expressed in mass and volume is rho = mass/volume or p = m/v. Rearranging the formula to isolate volume gives the formula v = m/p. To solve for the problem given, this formula must be used. This gives a solution of:

v = m/p = 250 g/ 968 g/cm^3 = 0.258 cm^3 of sodium

4 0

3 years ago