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3 years ago

A boy of mass 40kg while running develops a momentum of 180Ns. Calculate the velocity of the boy

Physics

1 answer:

Setler79 [48]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf v= 4.5 \ m/s}}$

Explanation:

The formula for momentum is:

$p=mv$

(p is momentum, m is mass, and v is velocity)

Let's rearrange the formula for velocity, or v.

$\frac{p}{m} =v$

Velocity can be found by dividing the momentum by the mass.

The momentum is 180 kilograms meter per second. The mass of the boy is 40 kilograms.

$p=180 \ kg*m/s \\m=40 kg$

Substitute the values into the formula.

$v=\frac{180 \ kg*m/s}{40 \ kg}$

Divide. Note that the kilograms or "kg" will cancel each other out.

$v=\frac{180 \ m/s}{40 }$

$v= 4.5 \ m/s$

The velocity of the boy is 4.5 meters per second.

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Leila is building an aluminum-roofed shed in her backyard to store her garden tools.The flat roof will measure 2.0 x 3.0m in are

IrinaK [193]

Answer: 0.0138 m^2 = 138 cm^2

Explanation:

The thermal expansion is the term use for the physical phenomena of dilation of the objects when they are exposed to changes in temperature.

The objects dilate when they are heated and contract when they are cooled.

The dilation is proportional to the change in temperatur.

For linear dilation, the proportionality constant is called linear dilation coefficient of the materials, it is named α and is measured in °C ^-1.

ΔL = α * Lo * ΔT, which means that the dilation (or contraction) is proportional to the product of the original length (Lo) and the change of temperature (ΔT).

There is also superficial dilation, for which the dilation is:

ΔA = β * Ao * ΔT, which means that the superficial dilation (or contraction) is proportional to the product of the original area (Ao) and the change of temperature (ΔT).

It is very interesting and important to solve problems that β = 2α, because regularly you will find the values of α for different materials and so, you just to multiply it times 2 to use β.

For this problem:

- Original area, Ao = area of the flat roof at - 10°C = 2.0m * 3.0m = 6.0 m^2.
- α for aluminum = 24 * 10^ -6 °C^-1.
- ΔT = 38°C - (-10°C) = 48°C

So, ΔA = 6.0m^2 * (2 * 24*10^-6 °C&-1) * 48°C = 0.0138 m^2

And that is the area that should stick out in summer to fit the structure during cold winter nights.

You can pass that number to cm^2 to grasp better the idea of this size:

0.0138 m^2 * (100 cm)^2 / m^2 = 138 cm^2

3 0

3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton

Brrunno [24]

Answer:

A) F_g = 4.05 10⁻⁴⁷ N, B) F_e = 9.2 10⁻⁸N, C) $\frac{F_e}{F_g}$ = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

F = $G \frac{m_1m_2}{r^2}$

let's calculate

F = $6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }$

F_g = 4.05 10⁻⁴⁷ N

B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

F = $k \ \frac{q_1q_2}{r^2}$

let's calculate

F = $9 \ 10^9 \ \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}$

F_e = 9.2 10⁻⁸N

C) It is asked to find the relationship between these forces

$\frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }$

= 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

4 0

3 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 25 m lower vert

Musya8 [376]

Answer:

x = 5.79 m

Explanation:

given,

mass of the car = 39000 Kg

spring constant = 5.7 x 10⁵ N/m

acceleration due to gravity = 9.8 m/s²

height of the track = 25 m

length of spring compressed = ?

using conservation of energy

potential energy is converted into spring energy

$m g h = \dfrac{1}{2}kx^2$

$x =\sqrt{\dfrac{2 m g h}{k}}$

$x =\sqrt{\dfrac{2\times 39000 \times 9.8 \times 25}{5.7 \times 10^{5}}}$

$x =\sqrt{33.5263}$

x = 5.79 m

the spring is compressed to x = 5.79 m to stop the car.

3 0

3 years ago

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

3 0

3 years ago

A boy of mass 40kg while running develops a momentum of 180Ns. Calculate the velocity of the boy​

A boy of mass 40kg while running develops a momentum of 180Ns. Calculate the velocity of the boy