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ludmilkaskok [199]

2 years ago

14

A ball is rolling across the floor at a constant speed. What will happen to the ball if it is exposed to an unbalanced force in

the same direction that it is moving?
A. It will immediately stop.
B. Its speed will increase.
C. Its speed will decrease.
D. Its motion will be unchanged.

2 answers:

KATRIN_1 [288]2 years ago

7 0

Answer:

B. it's speed will increase

lions [1.4K]2 years ago

4 0

Explanation:

B. Its speed will increase.

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A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring

Dennis_Churaev [7]

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

5 0

2 years ago

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)

where

m is the mass of the block

$v_{max}=1.25 m/s$ is the maximum speed

We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

$m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg$

b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

$k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m$

c)

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

#LearnwithBrainly

3 0

3 years ago

A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?

Vladimir [108]

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

$F=ma\\\\a=\frac{F}{m}$

where,

a = acceleration = ?
F = Magnitude of the applied force = 6 N
m = Mass of the ball = 9 kg

Therefore,

$a=\frac{6\ N}{9\ kg}$

a = 0.67 m/s²

Learn more about Newton's Second Law of Motion here:

brainly.com/question/13447525

#SPJ1

7 0

2 years ago

In creating his definition of horsepower, James Watt, the inventor of the steam engine, calculated the power output of a horse o

Studentka2010 [4]

Answer:

Part a)

$F = 182.3 Lb$

Part b)

$P = 0.55 HP$

Explanation:

Diameter of the circle = 24 ft

Diameter = 731.52 cm = 7.3152 m

now the horse complete 144 trips in one hour

so time to complete one trip is given as

$t = \frac{3600}{144} s$

$t = 25 s$

now the speed of the horse is given as

$v = \frac{2\pi r}{t}$

$v = \frac{\pi(7.3152)}{25}$

$v = 0.92 m/s$

Part a)

Now we know that the power is defined as rate of work done

it is given as

$P = F v$

$746 = F(0.92)$

$F = 810.9 N$

$F = 182.3 Lb$

Part b)

Work done to climb up to 3 m height is given by

$W = mgh$

now we have

$Power = \frac{Work}{time}$

$P = \frac{mgh}{t}$

$P = \frac{(70kg)(9.81)(3)}{5.0s}$

$P = 412.02 Watt$

now we know that 1 HP = 746 Watt

so we have

$P = \frac{412}{746} = 0.55 HP$

8 0

3 years ago

How are theories supported in the scientific community?

mafiozo [28]

They are unproven but accepted as fact.
Many experiments support them but they can be disproven by the results of a single experiment. Until then, they stand.
The third statement is correct.

7 0

3 years ago