Ask question

Ask question

All categories

2 years ago

8

katniss shoots an arrow horizontally frok a height of 1.8m. if the arrows initial velocity is 32m/s , how far will the arrow tra

vel before hitting the ground

1 answer:

Masteriza [31]2 years ago

3 0

Answer:

19.5 m

Explanation:

EN LANZAMIENTO DE PROYECTILES, LA VELOCIDAD HORIZONTAL

PERMANECE CONSTANTE.

PRIMERO ENCUENTRA LA VELOCIDAD VERTICAL

$v^{2}$ y = 2( $9.80 m/s^{2}$ )(1.8 m) = $\sqrt{35.3 m^{2}/s^{2}}$ = 5.94 m/s

LUEGO ENCUENTRAS EL TIEMPO QUE LA FLECHA PERMANECE EN EL

AIRE.

t = 5.94 m/s / $9.80 m/s^{2}$ = 0.61 s

FINALMENTE ENCUENTRAS LA DISTANCIA HORIZONTAL TOTAL.

x = 32 m/s(0.61 s) = 19.5 m

You might be interested in

A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere

Naya [18.7K]

Answer: a) B = 6811N

b) m = 603.2kg

c) 86.8%

Explanation: <em>Buoyant force</em> is a force a fluid exerts on a submerged object.

It can be calculated as:

$B=\rho_{fluid}.V_{obj}.g$

where:

$\rho_{fluid}$ is density of the fluid the object is in;

$V_{obj}$ is volume of the object;

g is acceleration due to gravity, is constant and equals 9.8m/s²

a) For the hollow plastic sphere, density of water is 1000kg/m³:

$B=10^{3}.0.695.9.8$

B = 6811N

b) Anchored to the bottom, the forces acting on the sphere are <u>Buoyant</u>, <u>Tension</u> and <u>Force due to gravity</u>:

B = T + $F_{g}$

B = T + mg

mg = B - T

$m=\frac{B - T}{g}$

Calculating:

$m=\frac{6811 - 900}{9.8}$

m = 603.2kg

c) When the shpere comes to rest on the surface of the water, there are only <u>buoyant</u> <u>and</u> <u>gravity</u> acting on it:

B = m.g

$\rho_{w}.V_{sub}.g=m.g$

$V_{sub}=\frac{m}{\rho_{w}}$

$V_{sub}=\frac{603.2}{1000}$

$V_{sub}$ = 0.6032m³

Fraction of the submerged volume is:

$\frac{V_{sub}}{V_{obj}}$ = $\frac{0.6932}{0.695}$ = 0.868

<u />

3 0

3 years ago

Hydrogen is a nonmetal with many nonmetal properties. Why is it on the periodic table with the metals in Group 1?

mrs_skeptik [129]

Answer:

This is because it has ns1 electron configuration like the alkali metals.

Pls Mark Brainiest, am trying to become Virtuoso

6 0

3 years ago

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. H

lukranit [14]

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2$

Let d is the distance moved in 2.25 s. Using second equation of motion,

$d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m$

So, it will move 6.32 m from rest in 2.25 seconds.

4 0

3 years ago

The world record for the 100.0 meter dash is 9.580 seconds. Find the speed of the runner in miles per hour.

grigory [225]

Answer:

233.1 miles per hours

Explanation:

Speed: This is defined as the ratio of distance to time. The S.I unit of speed is m/s. speed is a vector quantity because it can only be represented by magnitude only. Mathematically, speed can be expressed as,

S = d/t ....................... Equation 1

Where S = speed of the runner, d = distance covered, t = time.

Given: d = 100 meter , t = 9.580 seconds

Conversion:

If, 1 meter = 0.00062 miles

Then, 100 meters = (0.00062×100) miles = 0.62 miles.

Also

If, 3600 s = 1 h

Then, 9.580 s = (1×9.580)/3600 = 0.00266 hours.

Substitute into equation 1

S = 0.62/0.00266

S = 233.1 miles per hours.

Hence the runner speed is 233.1 miles per hours

7 0

3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago