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mina [271]
3 years ago
15

A wall lizard has adhesive pards in the limbs true or false plz help me​

Physics
1 answer:
Leto [7]3 years ago
5 0
True lizards can stick to surfaces because their bulbous toes are covered in hundreds of tiny microscopic hairs called setae
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A child tugs on a rope attached to a 0.62 kg toy with a horizontal force of 16.3 N. A puppy pulls the toy in the opposite direct
sattari [20]

Answer:

a=51.77\ m/s^2

Explanation:

Given that,

The mass of a toy, m = 0.62 kg

Force with which a child pull the toy = 16.3 N

The force with which the toy pulled in the opposite direction = -15.8 N

We need to find the acceleration of the toy. Let F be the net force acting on the toy. It is equal to :

F = 16.3 N - (-15.8 N)

= 32.1 N

Let a be the acceleration of the toy. Using Newton' second law of motion to find it.

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{32.1}{0.62}\\\\a=51.77\ m/s^2

So, the acceleration of the toy is 51.77\ m/s^2.

5 0
3 years ago
Which statement best explains acceleration?
Alexxx [7]
Acceleration is the ratio of a change in velocity  to the time over which the change happend
5 0
4 years ago
Read 2 more answers
A joule, which is a unit of work, is equal to?
sweet-ann [11.9K]
Option A is correct
4 0
3 years ago
Sean, after being so happy for two full days that he reported he "never needed much sleep," now is stating he is so sad that he
Kaylis [27]
D or A i'm not that sure
5 0
3 years ago
Read 2 more answers
One end of a metal rod is in contact with a thermal reservoir at 699. K, and the other end is in contact with a thermal reservoi
bulgar [2K]

Answer:

a)ΔS₁ = - 9.9 J/K

ΔS₂ = 69 J/K

b)The entropy change for the rod = 0 J/K

c)ΔS = 59.1 J/K

Explanation:

Given that

T₁ = 699 K

T₂= 101 K

Q= 6970 J

Change in entropy given as

\Delta S=\dfrac{Q}{T}

For 699 K:

\Delta S_1=\dfrac{Q}{T}

\Delta S_1=-\dfrac{6970}{699}

ΔS₁ = - 9.9 J/K  ( Negative because heat is leaving from the system)

For 101 K;

\Delta S_2=\dfrac{Q}{T}

\Delta S_2=\dfrac{6970}{101}

ΔS₂ = 69 J/K

The entropy change for the rod = 0 J/K

Entropy  change for the system

ΔS = ΔS₂  + ΔS₁

ΔS = 69 -9.9 J/K

ΔS = 59.1 J/K

8 0
3 years ago
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