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OleMash [197]
3 years ago
14

A 3,000-N force acts on a 200-kg object, what is the acceleration?​

Physics
2 answers:
Serjik [45]3 years ago
7 0

Answer:

15 m/s

Explanation:

For this problem we have to find the formula for acceleration.

F = ma

So since we already know force and looking for acceleration, we have to change it up a bit.

A = F / m

The force is 3000-N

A = 3000/ m

The mass is 200-kg

A = 3000/ 200

Do the division:

3000/ 200

15

A = 15

Zepler [3.9K]3 years ago
6 0

Answer:

15

Explanation:

a = F/ m

3000 / 200 = 15

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cluponka [151]

Answer:

P = 359.8 atm

Explanation:

The van der Waals' equation relates the properties of a gas, introducing constants "a" and "b" in order to consider gases as real gases. The equation is:

(P+a.\frac{n^{2} }{V^{2} } ).(V-nb)=n.R.T

where,

P: pressure

a: correction factor for intermolecular forces

V: volume

b: correction factor for molecules' volume

n: moles

R: ideal gas constant

T: absolute temperature

(P+\frac{1.390L^{2}atm}{mol^{2}}.\frac{(9.800mol)^{2}}{(0.8166L)^{2}}).(0.8166L-9.800mol.\frac{3.910 \times 10^{-2}L}{mol})=9.800mol \times \frac{0.08206atm.L}{mol.K} \times 301.8K\\(P + 200.2atm).(0.4334L) = 242.7atm.L\\P=359.8 atm

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3 years ago
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7 0
4 years ago
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The total number of protons and neutrons in an atom's nucleus is called 5
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6 0
3 years ago
Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?
Alina [70]
The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
F= k_e \frac{ q_1 q_2}{r^2}
where k_e=8.99\cdot10^9~N m^2 C^{-2} is the Coulomb's constant, q_1=q_2=29~nC=29\cdot 10^{-9}~C is the intensity of the two charges, and r=5.5~cm=0.055~m is the distance between them.

Substituting these numbers into the equation, we get
F=2.5~10^{-3}~N

The force is repulsive, because the charges have same sign and so they repel each other.
6 0
3 years ago
A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point
rjkz [21]

Answer:

Explanation:

given

initial velocity u = 4.45m/s

Height = 0.6m

g = 9.8m/s²

Required

final velocity v

Using the equation of motion;

v² = u²-2gH (upward motion of the fish makes g to be negative)

v² = 4.45²-2(9.8)(0.6)

v² = 19.8025-11.76

v² = 8.0425

v = 2.84 m/s

Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s

To get the time, we will use the formula

v = u - gt

2.84 = 4.45 - 9.8t

2.84-4.45 = -9.8t

-1.61 = -9.8t

t = 1.61/9.8

t = 0.164secs

Hence the time taken is 0.164secs

8 0
3 years ago
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