Kim took 5 hours to complete a journey with an average speed

A horizontal force acts on an object on a fric- tionless horizontal surface. If the force is halved and the mass of the object i

guapka [62]

(a) If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b) if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

$\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}$

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

$\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)$

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

$\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)$

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

3 0

3 years ago

The dimensions of the disk of our milky way galaxy are:.

Katen [24]

Answer:

According to studies, the milky way is approximately, "170,000–200,000 light-years (52–61 kpc) in diameter and, on average, approximately 1,000 ly (0.3 kpc) thick."

With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000

6 0

3 years ago

A fountain sends a stream of water straight up into the air to a maximum height of 5.60 m. The effective area of the pipe feedin

kifflom [539]

Answer:It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.

Read more on Brainly.com - brainly.com/question/1492542#readmore

Explanation:

5 0

3 years ago

A place kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear

trapecia [35]

<span>The ball clears by 11.79 meters Let's first determine the horizontal and vertical velocities of the ball. h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s Now determine how many seconds it will take for the ball to get to the goal. t = 36.0 m / 15.04 m/s = 2.394 s The height the ball will be at time T is h = vT - 1/2 A T^2 where h = height of ball v = initial vertical velocity T = time A = acceleration due to gravity So plugging into the formula the known values h = vT - 1/2 A T^2 h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2 h = 42.92 m - 4.9 m/s^2 * 5.731 s^2 h = 42.92 m - 28.0819 m h = 14.84 m Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>

4 0

4 years ago

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

Amanda [17]

Answer:

The tube should be held vertically and perpendicular to the ground.

Explanation:

Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:

Reasoning:

The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.

Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.

So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.

hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.

6 0

3 years ago