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Kay [80]
3 years ago
15

1. How much energy is needed to raise 1 g of water 1°C?

Physics
1 answer:
Serga [27]3 years ago
5 0

Answer:

1.The calorie was originally defined as the amount of heat required at a pressure of 1 standard atmosphere to raise the temperature of 1 gram of water 1° Celsius. Since 1925 this calorie has been defined in terms of the joule, the definition since 1948 being that one calorie is equal to approximately 4.2 joules.

2.Boiling water at 100 degrees Celsius: 540 calories are needed to turn 1 gram (at 100 degrees Celsius) of water to steam.

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Does anyone know this one? Thanks
Inessa [10]

Answer:

3.844\,*\,10^5

So a=3.844 and b=5

Explanation:

Scientific notation requests to write a number using powers of ten as a factor accompanying a real number (a) between 1 and smaller than 10 that contains the digits to exactly represent the original number. So in this case, the number 384,400 can be written as:

384,400=3.844 \,*\,100,000= 3.844 \,*\,10^5

with a=3.844, and "5" as the exponent of ten (so b=5)

6 0
4 years ago
Mass and Weight: What is the mass of an object that experiences a gravitational force of 685 N near Earth's surface where g
11Alexandr11 [23.1K]

Answer:

m = 69.9 kg

Explanation:

The mass and the weight of an object are two different quantities. Mass is basically the amount of matter that is present in a body. It remains same everywhere in the universe and measured in kilograms.

Weight is basically a force. It is the force by which earth attracts everything towards itself. The weight of an object changes from planet to planet, with the change in value of the gravitational acceleration (g).

Therefore, the relation between mass and weight of an object is given by the following formula:

W = mg

m = W/g

where,

m = mass = ?

W = Weight = 685 N

g = 9.8 m/s²

Therefore,

m = (685 N)/(9.8 m/s²)

<u>m = 69.9 kg</u>

4 0
3 years ago
Determine whether each of the statements below is true or false, and place it in the appropriate bin. Objects with equal speeds
lisov135 [29]

Objects with equal speeds definitely have equal velocities. -- FALSE.  For equal velocities, they also have to be going in the same direction.

If you are given an object's velocity, you can definitely determine its speed. -- TRUE.  If you know the velocity, then you know both the object's speed and its direction.

If you know the distance an object travels, and the time it takes to do so, you can determine the object's velocity. -- FALSE. Knowing the distance and time, you can figure out the object's speed.  But if you don't also know the direction it's moving, then you can't say what its velocity is.

If an object moves at constant speed, it must also be moving at constant velocity. -- FALSE.  Besides constant speed, it also needs to move in a straight line to have constant velocity.  If it turns, its velocity changes, even if its speed doesn't.

If an object moves at constant velocity, it must also be moving at constant speed. -- TRUE.  Constant velocity means its speed AND its direction are not changing.

Objects with equal velocities definitely have equal speeds. -- TRUE.  If their velocities are equal, then their speeds are equal AND they're moving in the same direction.

After laboring through this one, I'm wondering if there can possibly be any more ways to say the same thing.

7 0
3 years ago
Put the waves in order from shortest to longest wavelength
FromTheMoon [43]

Answer:

b, a, c

Explanation:

The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.

3 0
2 years ago
You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling a
Sonja [21]

Answer:

L=55.9m

Explanation:

The equation for the period of a simple pendulum is:

T=2\pi\sqrt{\frac{L}{g}}

In our case what we know is the period and the acceleration of gravity, and we need to know the length of the pendulum, so we can write:

L=(\frac{T}{2\pi})^2g

Which for our values is:

L=(\frac{15s}{2\pi})^2(9.81m/s^2)=55.9m

6 0
3 years ago
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