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bezimeni [28]
2 years ago
8

a horizontal force of 100N is required to push a crate across a factory floor at a constant speed. What is the net force acting

on the crate/ What is the force of friction acting on the crate?
Physics
1 answer:
Bas_tet [7]2 years ago
8 0

If the crate is moving along the floor in the same direction with a constant speed, it is in dynamic equilibrium. Equilibrium means there is no net force acting on the crate.

Since there is no net force on the crate, there must be a friction force on the crate equal in magnitude and opposite in direction to the applied horizontal force. Therefore the force of friction acting on the crate is 100N.

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A student starts a food fight by throwing a 0.5 kg burrito at some girl he likes. He throws it kind-of hard so he accelerates it
Elenna [48]

Explanation:

m = mass of burrito thrown by the student = 0.5 kg

a = acceleration of the burrito thrown by the student = 3 m/s²

F = force applied by the student on the burrito = ?

According to newton's second law , the net force on an object is the product of its mass and acceleration. it is given as

F = ma

inserting the values

F = (0.5) (3)

F = 1.5 N

hence the net force on the burrito comes out to be 1.5 N

4 0
2 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit
Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

d_1=4.7\times 10^{10} m

v_i=9.5\times 10^4 m/s

d_2=6\times 10^{12} m

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}

\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f

v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})

v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}

Using the formula

v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}

Where mass of sun=M=1.98\times 10^{30} kg

G=6.7\times 10^{-11}

v_f=58515.9 m/s

4 0
3 years ago
Consider a buggy being pulled by a horse.
rjkz [21]

Answer:

Option 2

Explanation:

8 0
3 years ago
When a puddle dries up what are the particles really doing
Alexandra [31]
The particles are either being absorbed or evaporating
7 0
2 years ago
Calculate (in MeV) the total binding energy for 40Ar. Express your answer in mega-electron volts to four significant figures.
AleksAgata [21]

Answer:

299.4 MeV

Explanation:

For 40 Ar

Number of neutrons = 22

Number of protons= 18

Mass of each proton = 1.007277 amu

Mass of each neutron= 1.008665 amu

Total mass of protons= 18 × 1.007277 amu = 18.130986 amu

Total mass of neutrons = 22 × 1.008665 amu = 22.19063 amu

Total mass of protons and neutrons= 18.130986 + 22.19063 = 40.321616 amu

Mass defect = 40.321616 amu - 40 amu

Mass defect = 0.321616 amu

Since 931 is the conversion factor from amu to MeV

Binding energy = 0.321616 amu × 931 = 299.4 MeV

Binding energy = 299.4 MeV

4 0
3 years ago
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