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3 years ago

A ransom note was found and the police need your help. They are backlogged with work and hopes your

Physics

1 answer:

Romashka-Z-Leto [24]3 years ago

3 0

Hello can someone please help me with this

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A 0.155 kg arrow is shot upward

Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0

3 years ago

Which voice can produce a pitch that has a speed of 343 m/s and a wavelength of 0.68 m?

lys-0071 [83]

The answer for you question would be : Soprano
One of the type of voice that could produce a pitch that has a speed of 343 m/s and a wavelength of 0.68 m would be soprano

hope this helps

8 0

3 years ago

Please help u guys acellus sucks

34kurt

Answer:

6.0cm

Explanation:

Given

focal length = 15.0cm

object distance = 10.0cm

Required

Image distance v

Using the formula

1/f = 1/u + 1/v

1/15 = 1/10+1/v

1/v = 1/15 + 1/10

1/v = 2+3/30

1/v = 5/30

v = 30/5

v = 6.0cm

Hence the image distance is 6.0cm

7 0

3 years ago

Is winter eye grass autotrophic

inessss [21]

Yes I do believe it is

6 0

4 years ago

Read 2 more answers

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago