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3 years ago

What is the net force if you pull a cart to the right with 5N of

Physics

1 answer:

postnew [5]3 years ago

4 0

Answer:

In this case the net force is zero.

According to Newton's third law of motion: For every action, there is an equal and opposite reaction. In this case the net force is zero.

Explanation:

Hope this helps!! :)

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A proton is ejected from the sun at a speed of 2 x 10^6 m/s. How long does it take for this proton to reach earth? Answer in hou

wolverine [178]

The distance from the Sun to the Earth is 149,600,000 km.

$d=149 600 000 \ km = 149 600 000 000 \ m = 1496 \cdot 10^8 \ m \\
v=2 \cdot 10^6 \ \frac{m}{s} \\ \\ \\
t=\frac{d}{v} \\ \\
t=\frac{1496 \cdot 10^8 \ m}{2 \cdot 10^6 \ \frac{m}{s}}=748 \cdot 10^2 \ s =\frac{748 \cdot 10^2 }{36 \cdot 10^2} \ h \approx \underline{\underline{20,78 \ h}}$

5 0

3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago

A box is pushed horizontally with constant speed across a rough horizontal surface. Which of the following must be true?

Pavel [41]

D

Because if an object is moving at a constant speed the force of friction must equal the applied (horizontal) force, and for it to be accelerating or decelerating, the force of friction and the applied force must be unequal

5 0

3 years ago

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m