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3 years ago

A car has a kinetic energy of 432,000j when traveling at 32 meters a second, what's the cars mass?

Physics

1 answer:

melomori [17]3 years ago

4 0

Answer:843.75kg

Explanation:

kinetic energy(ke)=432000j

Velocity(v)=32m/s

Mass(m)=?

Ke=(mxv^2)/2

432000=(mx32^2)/2

432000=(mx32x32)/2

Cross multiplying we get

432000x2=(mx1024)

864000=1024m

Divide both sides by 1024

864000/1024=1024m/1024

843.75=m

m=843.75kg

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Froghopper insects have a typical mass of around 12.5 mg and can jump to a height of 42.3 cm. The takeoff velocity is achieved a

Maksim231197 [3]

Answer:

2065.005 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.421\\\Rightarrow a=-\frac{0.421\times -9.81}{0.002}\\\Rightarrow a=2065.005\ m/s^2$

Acceleration of the frog is 2065.005 m/s²

6 0

2 years ago

A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at

storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

= 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A) angular frequency = ?

$f = \dfrac{v}{\lambda}$

$f = \dfrac{0.38}{0.187}$

$f =2.03\ Hz$

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

$K = \dfrac{2\pi}{\lambda}$

$K= \dfrac{2\pi}{0.187}$

$K =33.59\ m^{-1}$

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k x - ω t)

y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)

4 0

2 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with

Tanya [424]

Answer:

(a) $K_{small}=4839.3J$

(b) $K_{larger}=17406.4J$

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

$K=\frac{1}{2}Iw^2$

Where I is rotational of inertia of solid cylinder about its central axis.

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2$

Substitute the given values

$K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J$

For Part (b)

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2$

Substitute the given values

$K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J$

8 0

3 years ago

Long Snorkel: Inhalation of a breath occurs when the muscles surrounding the human lungs move to increase the volume of the lung

Mila [183]

Answer:

h = 0.362 m

Explanation:

The pressure equation with depth is

P₂ = $P_{atm}$ +ρ g h

The gauge pressure is

P2 - $P_{atm}$ = ρ g h

This is the pressure that muscles can create

P₂ - $P_{atm}$ = 3740 Pa

But still the person needs a small pressure for the transfer of gases, so

P₂ - $P_{atm}$ = 3740 - 188 = 3552 Pa

This is the maximum pressure difference, where the person can still breathe,

Let's clear the height

h = 3552 / ρ g

h = 3552 / (1000 9.8)

h = 0.362 m

This is the maximum depth where the person can still breathe normally.

4 0

3 years ago

Rank the following compounds in order of decreasing vapor pressure.

liubo4ka [24]

Hello ? What compounds are you referring to ?

3 0

3 years ago