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3 years ago

a 5.5 kg box is pushed across the lunch table.the net force applied to the box is 9.7 N.what is the acceleration of the box?

Physics

1 answer:

skelet666 [1.2K]3 years ago

5 0

Answer:

1.76m/s²

Explanation:

Acceleration is the time rate of change in velocity of a body. It is a vector quantity that is it has both magnitude and direction

From newton's second law of motion which states that the rate of change of momentum of a body is proportional to the applied force which takes place in the direction of force applied.

This law gives a formula which relate force, mass and acceleration.

Force = mass x acceleration

Given that force = 9.7N , mass = 5.5kg

Since force(F)= mass(m) x acceleration(a)

Therefore F = ma

Divide both sides by m

F/m = ma/m

Therefore,

Acceleration (a) = F/m

Acceleration = 9.7N ➗ 5.5kg

Acceleration = 1.76m/s²

The S. I unit of acceleration is m/s²

I hope this was helpful, Please mark as brainliest

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Match each term with the appropriate definition

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Answer:

opaque = 4

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3 years ago

Find the ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance betw

Snezhnost [94]

Answer:

The ratio of the new force over the original force is 16

Explanation:

Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:

$F_G=G\,\frac{M_1\,\,M_2}{D^2}$

So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:

$F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G$

which is 16 times the original force.

So the ratio of the new force over the original force is 16

5 0

3 years ago

A student has a weight of 655 N. While riding a roller coaster they seem to weigh 1.96x 103 N at the bottom of a dip that has a

Nadya [2.5K]

Answer:

The speed of the roller coaster at this point is 18.74 m/s.

Explanation:

Given that,

Weight of the student, W = 655 kg

Weight of the roller coaster, $F=1.96\times 10^3\ N$

Radius of the roller coaster, r = 18 m

At the bottom of the loop, the weight of the roller coaster us given by :

$F=W+\dfrac{mv^2}{r}$

If m is the mass of the roller coaster,

$W=mg$

$m=\dfrac{W}{g}$

$m=\dfrac{655}{9.8}$

m = 66.83 kg

So,

$F=W+\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{(F-W)r}{m}}$

$v=\sqrt{\dfrac{(1.96\times 10^3-655)\times 18}{66.83}}$

v = 18.74 m/s

So, the speed of the roller coaster at this point is 18.74 m/s. Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers

For a conclusion to be accepted by the scientific community, the data must be accurate, free from error, and

Setler79 [48]

The data must be reproducible: it means that the data will be reliable and representative and not that this data would lead to one conclusion and another data would lead to another.

Data being varied, unique or surprising does not have anything to do with acceptability of the conclusion.

8 0

3 years ago

Read 2 more answers

a 3.18-kg rock is released from rest at a height of 26.6 m. ignore air resistance and determine (a) the kinetic energy at 26.6 m

antiseptic1488 [7]

Answer:

(a) 0 J

(b) 828.96 J

(c) 828.96 J

(d) 828.96 J

(e) 0 J

(f) 828.96 J

Explanation:

Given:

Mass of the rock is, $m=3.18\ kg$

Initial height of the rock is, $h_{i}=26.6\ m$

Final height of the rock is, $h_f=0\ m$

Initial velocity of the rock is, $v_i=0\ m/s(Rest)$

Acceleration due to gravity is, $g=9.8\ m/s^2$

(a)

Initial kinetic energy is given as:

$K_i=\frac{1}{2}mv_i^2$

Plug in the given values and solve for ' $K_i$ '. This gives,

$K_i=\frac{1}{2}\times 3.18\times 0^2=0$

Therefore, initial kinetic energy is 0 J.

(b)

Initial potential energy is given as:

$U_i=mgh_i\\U_i=3.18\times 9.8\times 26.6=828.96\ J$

Therefore, initial potential energy is 828.96 J.

(c)

Mechanical energy is equal to the sum of kinetic and potential energy. It is always a constant value. Therefore,

$ME=K_i+P_i\\ME=0+828.96=828.96\ J$

Therefore, mechanical energy at 26.6 m is 828.96 J.

(d)

Now, at the final height of 0 m, the decrease in potential energy will be equal to the increase in the kinetic energy. In other words, the potential energy at the start will be converted to kinetic energy at the bottom.

Therefore, kinetic energy at the 0 m is given as:

$K_f=U_i=828.96\ J$

(e)

Potential energy at the final height is given as:

$P_f=mgh_f=3.18\times 9.8\times 0=0\ J$

Therefore, final potential energy is 0 J.

(f)

Total mechanical energy is a constant and doesn't depend on the height.

So, total mechanical energy at 0 m is same as that at 26.6 m.

Total mechanical energy is 828.96 J.

8 0

3 years ago