Business are necessary in order to preform which of the following functions

What two factors affect the rate of acceleration of an object?

Masja [62]

For help with this answer, we look to Newton's second law of motion:

Force = (mass) x (acceleration)

Since the question seems to focus on acceleration, let's get
'acceleration' all alone on one side of the equation, so we can
really see what's going on.

Here's the equation again:

Force = (mass) x (acceleration)

Divide each side by 'mass',
and we have: Acceleration = (force) / (mass) .

Now the answer jumps out at us: The rate of acceleration of an object
is determined by the object's mass and by the strength of the net force
acting on the object.

5 0

3 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

3 years ago

An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal

borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be 80cm.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is 20 cm.

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

#teamtrees #PAW (Plant And Water)

I hope this helps!

8 0

2 years ago

Read 2 more answers

PLS ANSWER FAST WILL GIVE BRAINLEST!!

egoroff_w [7]

Answer:

from

force =mass x acceleration

mass = force/acceleration

m = f/a