All categories

3 years ago

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

Physics

1 answer:

Amanda [17]3 years ago

4 0

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

You might be interested in

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

juin [17]

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same

7 0

3 years ago

Read 2 more answers

List the following in order from largest to smallest:

LiRa [457]

Organisms, organs, tissues, cell ....

a more specific answer would be:
organism, organ system, organ, tissue, cell, atom

4 0

3 years ago

Why is damage from sound waves is an issue on the launchpad but not in the air

elixir [45]

Because it is if you know you know and it is also helping the sentwcnde and air and confiscation

6 0

2 years ago

An athlete ran a distance of 100 meters in 9.83 Seconds<br> What’s his average speed

trasher [3.6K]

Answer:

10.1729399797mp/s

Explanation:

100m divided by 9.83s=

10.1729399797mp/s

6 0

3 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

1 year ago