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3 years ago

Find the net force and acceleration. 10 points. Will give brainliest.

Physics

2 answers:

Svetllana [295]3 years ago

5 0

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the FRICTION.

Force by Xavier = 25 N

Force of friction in the opposite direction of Xavier = f = umg

where, u = coefficient of kinetic friction.

so we here get as ,

Net force = 16 N

and acceleration = 2.13 m/s^2

Afina-wow [57]3 years ago

3 0

Answer:

a) 16 N

b) 2.13 m/s²

Explanation:

Draw a free body diagram of the tv stand. There are four forces:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force P pulling right.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

The net force in the x direction is:

∑F = P − Nμ

∑F = P − mgμ

∑F = 25 N − (7.5 kg) (10 m/s²) (0.12)

∑F = 16 N

Net force equals mass times acceleration:

∑F = ma

16 N = (7.5 kg) a

a = 2.13 m/s²

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Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

Len [333]

Answer:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

3 0

3 years ago

Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t

Law Incorporation [45]

Answer:

The block has an acceleration of $3 m/s^{2}$

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

$\sum F_{r} = ma$ (1)

Where $\sum F_{r}$ represents the net force, m is the mass and a is the acceleration.

$F_{x} + F{y} = ma$ (2)

The forces present in x are $F = 10 N$ and $f = 4 N$ (the friction force):

$F_{x} = 10 N - 4 N$

Notice that $f$ subtracts to $F$ since it is at the opposite direction.

$F_{x} = 6 N$

The forces present in y balance each other:

$F_{y} = 0$

Therefore:

$6 + 0 = ma$

$6 N = (2kg)a$ (3)

But $1 N = 1 Kg.m/s^{2}$ and writing (3) in terms of a it is get:

$a = \frac{6 Kg.m/s^{2}}{2 Kg}$

$a = 3 m/s^{2}$

So the block has an acceleration of $a = 3 m/s^{2}$ .

4 0

3 years ago

1. A woman driving in a car at 21 m/s pushes down on the gas pedal. 6 seconds later, she is moving at 36 m/s. What was her avera

Furkat [3]

a=Vf-Vi/t=36-21/6=15/6=2.5m/s^2

5 0

3 years ago

Explain why atoms only emit certain wavelengths of light when they are excited. Check all that apply. Check all that apply. Elec

JulsSmile [24]

Answer:

Explanation:

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Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. TRUE. Again, you can observe just the transition due the change of energy of an electron in the quantized energy level

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7 0

3 years ago