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2 years ago

A cyclist is travelling eastwards at a velocity

Physics

1 answer:

VMariaS [17]2 years ago

8 0

Answer:

0.02 m/s^2

Explanation:

change in velocity= 4.5m/s - 2.3m/s = 2.2 m/s

acceleration= change in velocity/change in time

acceleration= 2.2/120= 0.0183

= 0.02 (to 2 significant figures)

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A metre rule is used to measure a length. Which reading is shown to the nearest millimetre? A 0.7m B 0.76m C 0.761m D 0.7614m

Cloud [144]

Answer:

none

Explanation:

~both of them show to the nearest metre.

~millimeter has (mm) unit eg 0.7mm

4 0

3 years ago

Read 2 more answers

B. Projectile on cliff (range)

dimulka [17.4K]

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

cos 25 = v₀ₓ / v₀

sin 25 = Iv_{oy} / v₀

v₀ₓ = v₀ cos 25

v_{oy} = v₀ sin 25

v₀ₓ = 22 cos 25 = 19.94 m / s

v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

y = y₀ + v_{oy} t - ½ g t²

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

0 = 21 + 0.0192 t - ½ 9.81 t²

4.905 t² - 0.0192 t - 21 = 0

t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

t = $\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}$

t = $\frac{0.003914 \ \pm 4.13828}{2}$

t₁ = 2.07 s

t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

t = 2.07 s

now we can look up the distance traveled

x = v₀ₓ t

x = 19.94 2.07

x = 41.28 m

5 0

3 years ago

A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a

anastassius [24]

a) y(max) = 337.76 m

b) t₁ = 5.30 s the time for y maximum

c)t₂ = 13.60 s time for y = 0 time when the fly finish

d) vₓ = 30 m/s vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )

v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y = y₀ + v₀y*t - (1/2)*g*t² (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt = v₀y - g*t

dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g

v₀y = 60*sin60° = 60*√3/2 = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s the time for y maximum

And y maximum is obtained from the substitution of t₁ in equation (1)

y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max) = 337.76 m

Total time of flying (t₂) is when coordinate y = 0

y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for t₂

t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t = [51.96 ±√2700 + 3920]/9.8

t = [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)

t₂ = 13.60 s time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant

vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)

vy = 51.96 - 133.28 vy = - 81.32 m/s

The sign minus means that vy change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60 m

x = 408 m

5 0

2 years ago

You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$

height(h)=1 m

time taken=0.64 s

$h=ut+frac{at^2}{2}$

$1=0+\frac{a(0.64)^2}{2}$

$a=4.88 m/s^2and [tex]a=\alpha r$

where $\alpha$ is angular acceleration of pulley

$4.88=\alpha \times 5.2\times 10^{-2}$

$\alpha =93.84 rad/s^2$

And Tension in Rope

$T=m(g-a)$

$T=1.7\times (9.8-4.88)$

T=8.364 N

and Tension will provide Torque

$T\times r=I\cdot \alpha$

$8.364\times 5.2\times 10^{-2}=I\times 93.84$

$I=0.463\times 10^{-2} kg-m^2$

$I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2$

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0

3 years ago

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

$m = 25 * 12.5$

$m \approx 310$

6 0

3 years ago