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SCORPION-xisa [38]
3 years ago
15

An object is moving in the absence of a net force. Which of the following best describes the object’s motion? A. The object will

slow down at a constant rate until coming to rest B. The object will stop moving and remain at rest until acted on by a net force C. The object will continue to move at a constant speed but in a circular path D. The object will continue to move with a constant velocity
Physics
1 answer:
docker41 [41]3 years ago
3 0

Answer:

D. The object will continue to move with a constant velocity

Explanation:

According to Newton's first law also known as law of inertia, states that an object at rest will remain at rest or, if in motion, will remain in motion at constant velocity unless acted on by a net external force.

Therefore, An object moving in the absence of a net force will continue to move at a constant velocity

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How can two substances be made up of the same two elements but have very different properties?
asambeis [7]
So, basically, it might have different quantities, such as Carbon Dioxide (CO2) and Carbon Monoxide (CO), which are both made up of Carbon and Oxygen, but Carbon Monoxide has less oxygen. And they both have different properties, like CO is poisonous.
3 0
3 years ago
Read 2 more answers
A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
rusak2 [61]

Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

Force = 300 N

Find:

Spring constant in N / m

Computation:

Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

Spring constant in N / m = 6,000

8 0
3 years ago
Technician A says that the freezing point of a substance is the temperature at which the substance changes state from a solid to
pashok25 [27]

Answer:

Technician B is correct

Explanation:

Freezing is a method of conversion of substance in its liquid state to solid state. It is the process by which a liquid substance changes to a solid at a particular temperature.

Increasing the pressure and decreasing the temperature of a liquid increases its freezing point. For example, in other to freeze water i.e to change water to ice, it has to be kept in a fridge at a temperature lower than the temperature of the water. The essence of covering the fridge after placing the water in the fridge is to increase the pressure of the liquid hence increasing its freezing rate.

Based on the above explanation, it can be concluded that technician B is correct.

5 0
3 years ago
The Carson family's pancake recipe uses 2 teaspoons of baking powder for every 1/3 of a teaspoon of salt. How much baking powder
melomori [17]

Answer:

6 teaspoons of baking powder required.

Explanation:

Given that

According to the recipe of pancake,

For every \frac{1}{3} teaspoon of salt, 2 teaspoons of baking baking powder is required.

To find:

How much baking powder will be needed, if 1 teaspoon of salt was used ?

Solution:

This problem can be solved using ratio.

\frac{1}3 teaspoon of salt : 2 teaspoons of baking powder

Let us multiply the above ratio with 3.

\frac{1}{3}\times 3 teaspoon of salt : 2 \times 3 teaspoons of baking powder

OR

1 teaspoon of salt : 6 teaspoons of baking powder

So, answer is <em>6 teaspoons </em>of baking powder required.

Also, we can use the unitary method:

\frac{1}3 teaspoon of salt needs =  2 teaspoons of baking powder

1  teaspoon of salt needs =  \frac{2}{\frac{1}3} teaspoons of baking powder

1  teaspoon of salt needs = 2 \times 3 = <em>6</em>  teaspoons of baking powder needed

So, the answer is:

<em>6 teaspoons of baking powder </em>required.

5 0
3 years ago
Read 2 more answers
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

4 0
3 years ago
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