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2 years ago

Hi how are you?Answer only Philippines

Physics

2 answers:

Damm [24]2 years ago

5 0

Answer:

I'm just fine

Explanation:

Salamat sa Puntos

nataly862011 [7]2 years ago

4 0

Answer:

Mark me as brain list plz for fun

Explanation:

What do you mean??

And Mark me as brain list plz for i need to reach expert

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A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2

Akimi4 [234]

The acceleration exerted by the object of mass 10 kg is $\mathbf{1} m / \boldsymbol{s}^{2}$

Answer: Option A

<u>Explanation:</u>

According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

$Force = Mass \times Acceleration$

So,

$Acceleration=\frac{\text {Force}}{\text {Mass}}$

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

$Acceleration =\frac{10 \mathrm{N}}{10 \mathrm{kg}}=1 \mathrm{m} / \mathrm{s}^{2}$

So, the acceleration exerted by the object of mass 10 kg is $\mathbf{1} m / \boldsymbol{s}^{2}$

4 0

3 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

$m_3 = 0.59 kg$

3 0

3 years ago

Helppppppppppppppp.......

kow [346]

Answer:

9013 m/s

Explanation:

hope it helped!!!

8 0

3 years ago

Please help me to do this problem

Novosadov [1.4K]

Answer:

we got time and velocity over time.

so the distance is again the area underneath the graph

for a triangle with known base and height it's

4*10 / 2

distance traveled is 20

deceleration occurs when velocity decreases. that happens from t=2 till t=4

in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time

a (from t=2 to t=4) = -5v/t

7 0

3 years ago

An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached

Kruka [31]

The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring.

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be
<span>F = kx
270 N/m x 0.38 m = 102.6 N
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.

8 0

3 years ago

Hi how are you?Answer only Philippines​

Hi how are you?Answer only Philippines