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3 years ago

Identify the normal force on the shopping cart after 75 newtons of groceries are added to the cart

Physics

2 answers:

fgiga [73]3 years ago

7 0

Answer:

75 N upwards

Explanation:

The weight of the groceries acting vertically downward.

It works as teh action force acting on the ground.

Now according to the Newton's third law, for every action there is an equal and opposite reaction.

So, the reaction on the cart is same in magnitude but opposite in direction.

Action = weight of the groceries = 75 N acting downward

Reaction = Action but the direction is opposite

Normal reaction = 75 N upwards

Anna71 [15]3 years ago

3 0

If the shopping cart is parallel to the ground, the weight is acting vertically downward, then force is equal to weight. The normal force will be acting perpendicular to the ground,and acting upward opposite of the weight.

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El ancho de la uña de su dedo meñique se acerca a una unidad metrica comun ¿cual?

BartSMP [9]

Answer:

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3 0

3 years ago

A 30 kg student drops down from the monkey bars. The acceleration due to gravity is -9.8 m/s^2. Neglecting air drag, what is the

Sholpan [36]

The net force on the student is A) -294 N

Explanation:

Neglecting air resistance, there is only one force acting on the student: the force of gravity, which is given by

$F=mg$

where

m is the mass of the student

g is the acceleration of gravity

In this problem, we have:

m = 30 kg is the mass of the student

$g=-9.8 m/s^2$ is the acceleration of gravity, where the negative sign means the direction is downward

Substituting, we find the force of gravity on the student:

$F=(30)(-9.8)=-294 N$

And since this is the only force acting on the student, it is also the net force on him.

Learn more about gravitational force:

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8 0

3 years ago

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

Gravity at a height h above the surface of the moon will be given as:

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

3 years ago

Options are: a)4Cn b)5Cn c)6 Cn d)3 Cn

nasty-shy [4]

Answer:

Option B. 5 nC

Explanation:

From the question given above, the following data were obtained:

Capicitance (C) = 100 pF

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Next, we shall determine the quantity of charge. This can be obtained as follow:

Capicitance (C) = 1×10¯¹⁰ F

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Q = CV

Q = 1×10¯¹⁰ × 50

Q = 5×10¯⁹ C

Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:

1 C = 1×10⁹ nC

Therefore,

5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C

5×10¯⁹ C = 5 nC

Thus, the quantity of charge is 5 nC

3 0

2 years ago

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

3 years ago