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Simora [160]
2 years ago
15

A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push caus

es the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters?
Physics
1 answer:
Marina CMI [18]2 years ago
7 0

Answer:

v_{f} =4.9\frac{m}{s}  :  velocity of the crate after the person has pushed the crate a distance of 6 meters

Explanation:

Crate kinetics

Crate moves with uniformly accelerated movement

v f²=v₀²+2a*d (formula 1)

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Known data

v₀=0 The speed of the crate is equal to zero because part of the rest.

a= 2m/s²

d= 6m

Distance calculating

We replace data in the Formula (1)

v f²=0+2*2*d

v f²=2*2*6

v f²=24

v_{f} =\sqrt{24}

v_{f} =4.9\frac{m}{s}

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A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

We have the following data:

Speed of the wind from East to West: 6.68 m/s

Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

\theta=53.31\°

6 0
2 years ago
ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha
Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

4 0
2 years ago
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
Water is formed when two hydrogen atoms bond to an oxygen atom. The hydrogen and the oxygen in this example are different A) com
Alik [6]
C. Elements
elements are found in periodic table (in 1 box)
5 0
2 years ago
Read 2 more answers
If you observe an object in the universe that is roughly 8 megaparsecs in size, what is the object most likely to be?
Dmitrij [34]

E. Galaxy Cluster

Explanation:

A galaxy cluster, or cluster of galaxies, is a structure that consists of anywhere from hundreds to thousands of galaxies that are bound together by mutual gravity.

A megaparsec is a million parsecs and there are about 3.3 light years in a mega-parsec. Parsec is rather a natural distance unit for astronomers. The standard abbreviation of a mega-parsec is Mpc.

A parsec is approximately 3.09 x 1016 meters, a megaparsec is about 3.09 x 1022 meters.

Hence, 8 megaparsecs is gigantic size and that can be only of a galaxy cluster consisting of hundreds and thousands of galaxies bounded together.

Keywords: galaxies, parsec, megaparsec, galaxy cluster

Learn more about galaxy clusters and astronomical units from:

https://brainly.in/question/4624292

brainly.com/question/14214806

brainly.com/question/13315988

#learnwithBrainly

4 0
2 years ago
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