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Anestetic [448]
3 years ago
15

You walk for a speed of 1.1 m/s (meters per second) for 6.0 seconds and

Physics
1 answer:
castortr0y [4]3 years ago
5 0

Answer:

Average speed, As = 2.2 [m/s]

Explanation:

To solve these types of problems we must remember that the average of the speeds is determined by dividing the distance over time.

With the first speed and the time of 6 [s] we can calculate the distance.

V = x/t

where:

x = distance [m]

V = velocity = 1.1 [m/s]

t = time = 6 [s]

x1 = V*t

X1 = 1.1*6

X1 = 6.6 [m]

Now with the second velocity and  6 [s], we can calculate the second distance.

X2 = 3.3*6

X2 = 19.8 [m]

Now we have to calculate the average speed. The total distance is x = x1 +x2

X = 19.8 + 6.6 = 26.4 [m]

and the total time is 12 [s]

Therefore:

As = 26.4/12

As = 2.2 [m/s]

You might be interested in
How do Newton's laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can
lord [1]

Answer:

I'm not sure..but please refer to your teacher later.

Answer: Based on Newton's First law of motion (where inertia is involved), smooth ice increases the forceused to accelerate the hockey puck.

Explanation;

  • smooth ice reduces the resistances between the surface of the figure skates and the ice itself.
  • based on inertia theory ; the heavier the weight, the larger the inertia.. which explains it takes alot of force to move a heavier object than the lighter ones.. it also hard to *stop* the motion of heavier objects than the lighter ones.
  • now let's look at the design of the player shoe itself, they have a sharp blade at the bottom of the figure stakes.. which takes us to the law of the force.. the smaller the surface area, the more forces acting on it. So, players force (weight, F= mg) acts on the tip of the blade and on the ice
  • high inertia (run fast) and high force (attack opponent and pass puck) enables them to perform well in playing hockey
  • Thus if there's no resistance and the inertia of the player is high then they could run and pass the puck quickly
6 0
3 years ago
Read 2 more answers
Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us
nalin [4]

Answer:

13 W/m^2

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

I \propto \frac{1}{r^2}

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} (1)

where in this problem, we have:

I_1 = 1300 W/m^2 apparent brightness at a distance r_1, where

r_1 = 150 million km

We want to estimate the apparent brightness at r_2, where r_2 is ten times r_1, so

r_2 = 10 r_1

Re-arranging eq.(1), we find I_2:

I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2

5 0
3 years ago
Suppose you are standing at the exact center of a park surrounded by a circular road. An ambulance drives completely around this
vladimir2022 [97]

Answer: The pitch of the sound does not change.

Explanation: The pitch of sound wave is dependent on the frequency of the sound wave. The frequency of sound wave when their is a relative motion between an observer and a source is given by Doppler effect.

Doppler effect is a mathematical equation that gives the relationship between the observed frequency by an observer from any sound source and the relative motion between the observer and the source.

Mathematically,

f'= (v+v') /(v-vs) * f

Where f' = observed frequency of sound wave

v= speed of sound in air

v'= velocity of observer relative to sound source

vs= velocity of sound source relating to observer

f= frequency of sound produced by source

The observer is at the center, thus the distance between the observer and the source is constant (according to mensuration, the radius is constant for any given circle and since the car is moving along a circular path and the observer is at the center, thus the distance between them is constant), thus making the relative velocity between the observer and the source constant (vs=constant).

Also the frequency of sound wave produce by the source is a constant (f=constant)

The speed of sound in air is also a constant (v=336m/s)

The observer is standing at the center thus he is not moving, hence the relative motion between observer and source is also constant (v'=constant)

Since all parameters are constant, then the observed frequency will be constant too.



4 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
If two is company and three is a crowd, what is 4 and 5?
LiRa [457]

4 and 5 is 9

(4 + 5)

4 0
3 years ago
Read 2 more answers
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