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Pachacha [2.7K]
3 years ago
8

A force F produces an acceleration a on an object of mass m. A force 3F is exerted on a second object, and an acceleration 8a re

sults. What is the mass of the second object in terms of m?
a. 3m
b. 9m
c. 24m
d. (3/8)m
e. (8/3)m
Physics
1 answer:
Pie3 years ago
3 0

To solve this problem we will proceed to use Newton's second law for which the mass (m) multiplied by the acceleration (a) is defined as the Force (F) applied on a body, mathematically that is,

F = ma

According to the statement for the first object, the acceleration is,

a = \frac{F}{m} \rightarrow 1^{st} Object

For the second object the acceleration is,

8a = \frac{3F}{m_2} \rightarrow 2^{nd} Object

Solving for the mass of the second object,

m_2 = \frac{3F}{8a}

m_2 = \frac{3F}{8(F/m)}

m_2 = \frac{3}{8}m

Therefore the correct answer is D.

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Un cuerpo se mueve con MRU a una rapidez de 12 m/s y recorre 240 m; luego, disminuye su ra¬pidez uniformemente a razón de 2 m/s2
Annette [7]

Answer:

29 seconds

Explanation:

First we have a constant speed of 12 m/s and the distance of 240 m, so to find the time we can use the formula:

distance = speed * time

240 = 12 * time1

time1 = 20 seconds

Then, the speed decreases at 2 m/s2 until it reaches 2 m/s. So to find this time, we use this formula:

Final speed = inicial speed + acceleration * time

2 = 12 - 2 * time2

2*time2 = 10

time2 = 5 seconds.

Then, the speed increases from 2 m/s to 22 m/s with an acceleration of 5 m/s2, so we have:

Final speed = inicial speed + acceleration * time

22 = 2 + 5 * time3

5*time3= 20

time3 = 4 seconds

The total time is:

Total time = time1 + time2 + time3 = 20 + 5 + 4 = 29 seconds

7 0
3 years ago
A 2 feet piece of wire is cut into two pieces and once a piece is bent into a square and the other is bent into an equilateral t
SpyIntel [72]

Answer:

Explanation:

Let x ft be used to make square and 2-x ft be used to make equilateral triangle.

each side of square = x/4

area of square = ( x /4 )²

Each side of triangle

= (2-x) /3

Area of triangle = 1/2 (2-x)²/9 sin 60

= √3 / 36 x (2-x)²

Total area

A = ( x /4 )² +√3 / 36 (2-x)²

For maximum area

dA/dx = 0

1/16( 2x ) -√3 / 36 x2(2-x) = 0

x / 8 - √3(2-x)/ 18 = 0

x / 8 - √3/9 + √3/18 x = 0

x ( 1/8 + √3/18 ) = √3/9

x(.125 +.096 ) = .192

x = .868 ft

3 0
2 years ago
You pull a rope to the left with 300 N and a friend pulls the rope to the right with 425 N
eimsori [14]
I’m assuming you’re supposed to calculate the resultant force?

425N (right) -300N (left)
=125 N to the right
6 0
3 years ago
Hi If the coefficient of kinetic friction between the 5.0 kg mass and the table is 0.305, what is the tension in the string?
statuscvo [17]

Answer:

18 N

Explanation:

Draw a free body diagram for each block.

There are four forces acting on block I:

Weight force Mg pulling down

Normal force N pushing up

Tension force T pulling right

Friction force Nμ

There are two forces acting on block II:

Weight force mg pulling down

Tension force T pulling up

Sum of forces on block I in the +y direction:

∑F = ma

N − Mg = 0

N = Mg

Sum of forces on block I in the +x direction:

∑F = ma

T − Nμ = Ma

T − Mgμ = Ma

Sum of forces on block II in the -y direction:

∑F = ma

mg − T = ma

Solve for a in the first equation, then substitute into the second.

a = (T − Mgμ) / M

mg − T = m (T − Mgμ) / M

mMg − MT = mT − mMgμ

mMg + mMgμ = mT + MT

mMg (1 + μ) = (m + M) T

T = mMg (1 + μ) / (m + M)

T = (2) (5) (9.8) (1 + 0.305) / (2 + 5)

T = 18.27

Rounding to two significant figures, the tension is 18 N.

4 0
3 years ago
A 25.0-kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0°
SCORPION-xisa [38]

(a) 139.7 J

The potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=mg\Delta h

where

m = 25.0 kg is the mass of the child

g = 9.8 m/s^2

\Delta h is the difference in height between the initial position and the bottom position

We are told that the rope is L = 2.20 m long and inclined at 42.0° from the vertical: therefore, \Delta h is given by

\Delta h = L - L cos \theta =L(1-cos \theta)=(2.20 m)(1-cos 42.0^{\circ})=0.57 m

So, her potential energy is

U=(25.0 kg)(9.8 m/s^2)(0.57 m)=139.7 J

(b) 3.3 m/s

At the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

U=K=\frac{1}{2}mv^2

where

m = 25.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(139.7 J)}{25.0 kg}}=3.3 m/s

(c) 0

The work done by the tension in the rope is given by:

W=Td cos \theta

where

T is the tension

d is the displacement of the child

\theta is the angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. Therefore, \theta=90^{\circ} and cos \theta=0, so the work done is zero.

7 0
3 years ago
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