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Naddika [18.5K]
3 years ago
15

How’d the turtle cross the road

Physics
2 answers:
Mumz [18]3 years ago
7 0

Answer:

He crawled.

Explanation: He crawled with the strength he gained from a leaf.

cestrela7 [59]3 years ago
4 0

He did it by doing it.....

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What two factors determine how fast weathering occurs
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<span>The factors that determine how fast weathering occurs are the type of rock, type of soil, time it takes, and the climate.</span>
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The orientation of which of the following does not influence the phases of the moon? a. Earth c. Sun b. the moon d. Stars Please
beks73 [17]
I'm pretty sure it's D. The stars don't influence the moon's phases.
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3 years ago
1. What is the total distance traveled?<br> A 3.0m <br> B 4.0m <br> C 5.0m <br> D 6.0m
Ostrovityanka [42]

Answer:

c

Explanation:

7 0
3 years ago
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
4 years ago
The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold
Liula [17]

Answer:

L_{o}=0.1224m

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m

6 0
4 years ago
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