When people don't clearly recognize the obstacle causing their frustration, they

1. An airow is shot horizontally from a crossbow 1.5 m above the ground. The initial velocity is 45

Amanda [17]

Answer:

Time taken by the arrow to travel along to hit the ground is 0.55 seconds.

Explanation:

The only "force" acting on the "crossbow" to cause it to "hit" the ground is "gravity". There is no initial velocity downward when it shoot.

$d=v_{i} t+\frac{1}{2} t^{2}$

d = the displacement of the object

t = the time for which the object moved

a = acceleration of the object

$v_i$ = the initial velocity of the object

Given values

d = 1.5 m

t = unknown

$a=g=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mathrm{V}_{\mathrm{i}}=0 \mathrm{m} / \mathrm{s}$

$1.5 \mathrm{m}=0(\mathrm{t})+\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \mathrm{t}^{2}$

$1.5=0+4.9 \mathrm{t}^{2}$

$\mathrm{t}^{2}=\frac{1.5}{4.9}$

$t^{2}=0.306 \mathrm{s}$

Square root both sides

$t=\sqrt{0.306}$

t = 0.55 s

4 0

3 years ago

A spherical balloon has a radius of 8.35 m and is filled with helium. how large a cargo can it lift, assuming that the skin and

rusak2 [61]

Assuming that the densities of the gases are:

density of air, ρ1 = 1.29 kg / m^3

density of helium, ρ2 = 0.179 kg / m^3

Since buoyant force and weight are two forces that are in opposite direction (buoyant force is up while weight is down), therefore equate the two:

buoyant force = weight

m g = (800 + m1) g

where m is the mass of buoyancy, g is gravity and m1 is the maximum mass of the cargo

m = 800 + m1

We know that mass is also expressed as:

m = ρ V

where ρ is density of gas and V is volume of the sphere

Since there are two interacting gases here, therefore m is:

m = (ρ1 – ρ2) V

Therefore:

(ρ1 – ρ2) V = 800 + m1

(1.29 – 0.179) (4π/3) (8.35m)^3 = 800 + m1

2709.33 = 800 + m1

m1 = 1,909.33 kg

7 0

2 years ago

A worker on the roof of a house drops his 0.46 kg hammer, which slides down the roof at constant speed of 9.88 m/s. The roof mak

nekit [7.7K]

Answer:

The horizontal distance travelled in that time lapse is 12.94 m

Explanation:

In order to solve this problem, we'll need:

The horizontal speed
the time the hammer takes to fall from the roof to the ground

At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

$v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s$

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates

$x(t)=8.8 m/s *t$

$y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}$

Now we calculate the time the hammer takes to get to the floor

$17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s$ or $t=-2.38s$

Now, we keep the positive time result and calculate the horizontal distance travelled

$x(1.47s)=8.8m/s*1.47s=12.94m$

3 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

2 years ago

Which example identifies a change in motion that produces acceleration?

kifflom [539]

Acceleration is the rate at which an object changes its velocity. It defines how much the velocity is changing. The acceleration can be negative and positive. Negative acceleration is when the object slows down, while positive while the object goes faster.
<span>A ball moving at a constant speed around a circular track produces acceleration. </span>

6 0

2 years ago

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