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Andrews [41]
2 years ago
5

What is meant by error​

Physics
1 answer:
Romashka [77]2 years ago
3 0

Answer:

An error (from the Latin error, meaning "wandering") is an action which is inaccurate or incorrect. In some usages, an error is synonymous with a mistake. In statistics, "error" refers to the difference between the value which has been computed and the correct value.

hope it will help you

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The diagram below shows a boat moving north in a river at 3m/s while the current in the river moves south at 1 m/s.
salantis [7]

Answer it will move slower:

Explanation: cause the force that’s opposite of it

4 0
1 year ago
a girl of mass 40 kg wearing high heel of radius 3mm. find the pressure exerted by her on the ground. [answer the question fast]
kakasveta [241]
You should calculate 40 kg and the radius 3mm.
7 0
3 years ago
A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00
den301095 [7]

Answer:

\theta=7^o

Explanation:

<u>Displacement</u>

It is a vector that points to the final point where an object traveled from its starting point. If the object traveled to several points, then the individual displacements must be added as vectors.

The mail carrier leaves the post office and drives 2 km due north. The first displacement vector is

\vec r_1=\ km

Then the carrier drives 7 km in 60° south of east. The displacement has two components in the x and y axis given by

\vec r_2=\ km=\ km

Finally, he drives 9.5 km 35° north of east.

\vec r_3=\ km=\ km

The total displacement is

\vec r_t=\ km+\ km+\ km

\vec r_t=\ km

The direction can be calculated with

\displaystyle tan\theta=\frac{1.39}{11.28}=0.1232

\boxed{\theta=7^o}

7 0
3 years ago
In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move
lara31 [8.8K]

Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.  

The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.

8 0
2 years ago
The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104
Mamont248 [21]

To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

Centripetal acceleration can be found through the relationship

a_c = \frac{v^2}{R}

Where

v = Tangential Velocity

R = Radius

At the same time linear velocity can be expressed in terms of angular velocity as

v = R\omega

Where,

R = Radius

\omega = Angular Velocity

PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as

\omega = \frac{2\pi}{T}

T = Period

So replacing we have to

\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s

Since 1 Light year = 9.48*10^{15}m

Then the radius in meters would be

R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})

R = 2.844*10^{20}m

Then the centripetal acceleration would be

a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2

From the result obtained, considering that it is an unimaginably low value of an order of less than 10^{-10} it is possible to conclude that it supports the assertion on the inertial reference frame.

8 0
3 years ago
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