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2 years ago

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A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the

mass of the bike?

1 answer:

UkoKoshka [18]2 years ago

3 0

Answer:

Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, $KE = \frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

$1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg$

So, mass of bike = 38 kg.

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The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61

nikitadnepr [17]

Answer:

(a) Angle of incidence for violet is more than the angle of incidence for red

(b) 2.4°

Explanation:

refractive index for violet , v = 1.66

refractive index for red, nR = 1.61

wavelength for violet, λv = 400 nm

wavelength for red, λR = 700 nm

Angle of refraction, r = 30°

(a) Let iv be the angle of incidence for violet.

Use Snell,s law

nv = Sin iv / Sin r

1.66 = Sin iv / Sin 30

Sin iv = 0.83

iv = 56°

Use Snell's law for red

nR = Sin iR / Sin r

where, iR be the angle of incidence for red

1.61 = Sin iR / Sin 30

Sin iR = 0.805

iR = 53.6°

So, the angle of incidence for violet is more than red.

(b) iv - iR = 56° - 53.6° = 2.4°

4 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago

Tammy leaves the office, drives 26 km due

fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0

2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

8 0

2 years ago