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iragen [17]
3 years ago
15

A 39.7 n object is in free fall. what is the magnitude of the net force which acts on the object? answer in units of n.

Physics
1 answer:
Vika [28.1K]3 years ago
6 0
If there is no air resistance, the value would still be the same. So therefore, the answer is 39.7 N.
Supposing that there is an air resistance of 19.8 N.With a 19.8 N air resistance: ∑F = 39.7 N - 19.8 N = 19.9 N would be the net force. The important point here is that the force of gravity equivalent the weight of the object.
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A student placed a ladder up against a wall as shown below. The normal force applied by the wall in the ladder will be directed:
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The normal force is always perpendicular to the surface. So it would be straight to the left of the wall
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3 years ago
A rocket has a mass 280(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.
saw5 [17]

To solve this problem we will apply the concepts related to the conversion of units for which we will have that 1 slug is equal to 14.59kg. At the same time we will use Newton's second law for which weight is defined as the product between mass and acceleration (Due to gravity). This is then

A: Using the conversion ratio of slug to kilogram we have to,

1 slug = 14.59kg

Then

m = 280*10^3 slugs (\frac{14.59kg}{1slug})

m = 4.09*10^6kg

B: Using Newton's second law we have to,

W = mg

W = (4.09*10^6)(9.8)

W = 40034960N\approx 4*10^7N

8 0
3 years ago
9. In a closed circuit made of a battery, a wire, a lightbulb, and a switch which part's main responsibility is allowing electri
Zielflug [23.3K]

Answer:

d

Explanation:

6 0
2 years ago
Read 2 more answers
A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge
zheka24 [161]

Answer:

The acceleration of an electron is 1.2\times10^{20}\ m/s^2

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

F_{1}=\dfrac{kq_{1}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}

F_{1}=8\times10^{-11}j\ \ N

We need to calculate the force on electron due to q₂

Using formula of force

F_{2}=\dfrac{kq_{2}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}

F_{2}=6.93\times10^{-11}i\ \ N

We need to calculate the net force

Using formula of net force

F=F_{1}+F_{2}

Put the value into the formula

F=8\times10^{-11}j+6.93\times10^{-11}i

The magnitude of the net force

F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}

F=1.058\times10^{-10}\ N

We need to calculate the acceleration of an electron

Using formula of force

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}

a=1.2\times10^{20}\ m/s^2

Hence, The acceleration of an electron is 1.2\times10^{20}\ m/s^2

3 0
3 years ago
During a soccer game, a player grabs and holds an opponent's shirt outside of the penalty box. After the foul is called, what ki
Mila [183]
D direct free kick
Hope this helps
8 0
2 years ago
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