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Ivahew [28]
2 years ago
7

Andrew and Warren sit on opposite sides of a table of mass 28 kg. Andrew pushes the table to the right with a force of 277 N, an

d Warren pushes the table to the left with a force of 215 N. (Assume there is no friction.)
a. Draw the free-body diagram of the table. (You do not need to draw it perfectly to scale. Just make sure the directions are correct.)

b. Write the expression for the net force on the table along the y-axis.

c. Write the expression for the net force on the table along the x-axis.

d. What is the normal force acting on the table?

e. What is the net force on the table along the x-axis, including direction?

f. What is the acceleration of the table, including direction?
Physics
1 answer:
Ostrovityanka [42]2 years ago
4 0

The acceleration is 2.2 m/s^2 and the net force is  62N towards the right

<h3>What is the net force?</h3>

The net force is the effective force that acts on a body in a give direction.

1. The net force in the y axis is; ∑Fy = 0

2. The net force along the y axis is; ∑Fx =  277 N - 215 N

3. The normal reaction is given by;  28 kg * 9.8 m/s^2 = 274.4 N

4. The net force in the x axis is  277 N - 215 N = 62N towards the right

5. The acceleration of this force is obtained from;

F = ma

62 = 28 a

a = 62/28

a = 2.2 m/s^2

Learn more about the net force:brainly.com/question/18031889

#SPJ1

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Alex_Xolod [135]

Answer:

-4.5 m/s

Explanation:

Even without doing calculations, we know that a projectile has the same speed coming down as it did going up at the same height.  Since it went up at 4.5 m/s, it returns down at -4.5 m/s.

To prove it with math, use the given information:

v₀ = 4.5 m/s

a = -9.8 m/s²

Δy = 0 m

Find: v

v² = v₀² + 2aΔy

v² = (4.5 m/s)² + 2 (-9.8 m/s²) (0 m)

v = -4.5 m/s

3 0
3 years ago
A 6 V battery is connected to a 24 ohm resistor to create a circuit. The 6 V battery is then replaced with a 12 V battery. How d
11Alexandr11 [23.1K]

Answer:

1.) The current is doubled

2.) moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

Explanation:

1.) When 6 V battery is connected to a 24 ohm resistor to create a circuit, using ohms law, V = IR

Current I = 6/24 = 0.25 A

When the The 6 V battery is replaced with a 12 V battery,

Current I = 12/24 = 0.5 A

Therefore, The current is doubled

2.) Electric current will be induced when moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

The force of attraction between two different masses is inversely proportional to the square of the distance between them.

6 0
4 years ago
Two protons are 1 × 10−10 m apart (about one atomic radius). Which interaction between two protons is stronger, the gravitationa
anastassius [24]

Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.

Explanation: Please see the attachments below

4 0
4 years ago
Your friend says that for a moving object to continue moving, a force must be continually applied to it. Do you agree with your
Zigmanuir [339]

Answer:

I would disagree with my friend because that according to Newton's first law, an object in motion will continue to be in motion until stopped by another object/force. If the object is already moving, it will stay in motion until something else stops it. There is no need for a force to be <em>continually</em> applied to it while it's already moving.

4 0
2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
3 years ago
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