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Helga [31]
2 years ago
7

What is the difference between panda and rhino?​

Physics
2 answers:
olya-2409 [2.1K]2 years ago
7 0

Answer:

Explanation:

The difference between panda and rhino are :

<u>panda</u>

1. The giant panda, also known as the panda bear, is a bear native to South Central China

2. Their thick black and white fur equips them for life in cool forests

<u>Rhino</u>

1. There are 5 species of rhino.

2. Rhinos have poor vision

Jlenok [28]2 years ago
4 0

Answer:

plz mark me as brainliest

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What is a change in the speed of an object called?
mina [271]

Acceleration........................................

5 0
3 years ago
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A carpenter lifts a 10-kg piece of wood to his shoulder 1.5 m above the ground. He then sets the wood on his truck at 1.0 m abov
FinnZ [79.3K]

Answer:

<em>500Joules</em>

Explanation:

Kinetic energy = 1/2mv²

m is the mass of the wood

v is the velocity

Given

Mass = 10kg

Velocity v = 10m/s

Substitute into the formula and get KE

KE = 1/2 * 10 * 10²

KE  = 1/2 * 1000

KE = 500Joules

<em>Hence the kinetic energy of the wood during delivery is 500Joules</em>

8 0
2 years ago
The distance between Earth and Mars is 225 million km. When converted using the conversion factor 1 AU = 1.5 × 108 km, the dista
noname [10]
To convert km to AU, we divide 225,000,000 km by the factor of 1.5 x 10^8 = 150,000,000 km. This gives us 225,000,000 / 150,000,000 = 1.5 AU. Therefore, the distance between Earth and Mars in AU is 1.5 AU.
The AU is not equivalent to a light-year. A light-year is equivalent to around 9.5 x 10^12 kilometers.

4 0
3 years ago
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0
3 years ago
What force must the worker exert to get the box moving &amp; what force must the worker exert to accelerate the box at 0.1 meter
photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

Now if the box is already moving then we will have to use kinetic friction force between box and floor

now we can write the equation of net force as

F - F_k = m*a

here

F_k = kinetic friction = 220 N

m = mass = 500 kg

a = acceleration = 0.1 m/s^2

now we will have

F - 220 = 500* 0.1

F = 220 + 50 = 270 N

3 0
2 years ago
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