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viktelen [127]
3 years ago
6

Please match the correct letter to the scenario. ***********Worth 50 POINTS*******WILL MARK BRAINLIEST

Physics
1 answer:
MissTica3 years ago
7 0

Answer:1=C 2=A 3=B 4=E 5=G 6=G 7=A 8=E 9=C 10=A

Explanation:

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de
kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0
3 years ago
Explain the meaning of the error​
joja [24]

Answer:

a mistake

Explanation:

7 0
2 years ago
A car has a mass of 1000 kg. What is the acceleration produced by a force of 2000 N?
EastWind [94]
F=ma
a=F/m
a=2000/1000
a=2 m/s^2
6 0
2 years ago
On a cross-country trip, a couple drives 500 mi in 10 h on the first day, 380 mi in 8.0 h on the second day, and 600 mi in 15 h
Zepler [3.9K]

We know that the average speed is simply the ratio of the total distance travelled over the total duration of the trip.

total distance = 500 mi + 380 mi + 600 mi

total distance = 1,480 mi

 

total time = 10 h + 8 h + 15 h

total time = 33 h

 

So the average speed is therefore:

average speed = 1,480 mi / 33 h

<span>average speed = 44.85 mi / h</span>

8 0
3 years ago
What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe
Eduardwww [97]

Answer:

Final temperature, T_f=21.85^{\circ}

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, T_i=41^{\circ}C

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, c=0.235\ J/g\ C

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{-Q}{mc}+T_i

T_f=\dfrac{-18}{4\times 0.235}+41

T_f=21.85^{\circ}

So, the final temperature of silver is 21.85 degrees Celsius.

5 0
3 years ago
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