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3 years ago

11

Chi walked 100 meters in 5 minutes. She took another 15 minutes to walk another 400 meters to reach her house. What is her avera

ge speed?

1 answer:

Law Incorporation [45]3 years ago

4 0

Average speed is the ratio of total distance moved by Chi in total time interval

So here we will have

Total distance = 100 m + 400 m

$d = 500 m = 0.5 km$

Total time taken = 5 min + 15 min = 20 min

$T = \frac{20}{60} = \frac{1}{3} hour$

now by the formula of average speed we know that

$v_{avg} = \frac{d}{T}$

$v_{avg} = \frac{0.5 km}{1/3 h}$

$v_{avg} = 1.5 km/h$

so average speed will be 1.5 km/h

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Kipish [7]

Answer:

Velocity

Explanation:

"The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s."

^^This explanation is from physicsclassroom.com

3 0

3 years ago

What is the IMA of the following pulley system?<br><br>34567

Lynna [10]

Answer:

IMA of given system = $\frac{F_{r} }{F_{e} }$

Explanation:

The "Ideal Mechanical advantage" (IMA) of given pulley is $\frac{F_{r} }{F_{e} }$ .

Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by $F_{r}$ . The input force is analogous or equivalent to the effort applied i.e. $F_{e}$ .

Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is $\frac{F_{r} }{F_{e} }$ .
Its mathematical reference would be:

IMA = $\frac{F_{r} }{F_{e} }$

6 0

3 years ago

Where does the cart have the most velocity(speed)?

Varvara68 [4.7K]

Answer:

point c

Explanation:

the cart has accelerated and is at the lowest point on the path .

consider the acceleration due to gravity converting potential to kinetic energy

6 0

3 years ago

Read 2 more answers

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0

3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago