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3 years ago

14

A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha

t is the displacement?

1 answer:

kifflom [539]3 years ago

3 0

Answer:

$x_{distance}=10.07km$

Explanation:

Given data

time=0.530 h

Average velocity Vavg=19.0 km/s

To find

Displacement Δx

Solution

The Formula for average velocity is given as

$V_{avg}=(x_{distance} )/(t_{time} )\\ x_{distance}=V_{avg}*(t_{time} )\\x_{distance}=(19.0km/h)*(0.530h)\\x_{distance}=10.07km$

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Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.

LuckyWell [14K]

Answer:

I'm pretty sure the answer is 0 m/s²

Explanation:

The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.

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2 years ago

A man pushes his lawnmower with a velocity of +0.75 m/s relative to the ground. A girl rides by on her bike with a velocity of +

SashulF [63]

Answer:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are $V_a$ and $V_b$ respectively, the relative velocity of B with respect to A is given by:

$V_{ba}=V_b-V_a$

Thus, the velocity of the girl relative to the lawnmower is:

$V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}$

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2 years ago

Read 2 more answers

A 2.0-kilogram ball rolls down a ramp. if the ball accelerates at a rate of 12 m/s2, the net force causing the acceleration is:

ehidna [41]

Answer:

D

Explanation:

f = ma

2 x 12 = 24

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1 year ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

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2 years ago

The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length

Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

$y_r=1.22\frac{\lambda f}{d}$

Where,

$y_r =$ Range of the radius

$\lambda =$ wavelength

f= focal length

Our values are given by,

State 1:

$d=2.00mm = 2*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 750nm = 750*10^{-9}m$

State 2:

$d=8.00mm = 8*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 390nm = 390*10^{-9}$

Replacing in the first equation we have:

$y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}$

$y_{r1}= 11.4\mu m$

And also for,

$y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}$

$y_{r2} = 1.49\mu m$

Therefor, the airy disk radius ranges from $1.49\mu m$ to $11.4\mu m$

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2 years ago