A block of mass m 5.00 kg ispulled along a horizontal frictionless floor by a cord
that exerts a force of magnitude F 12.0 N at an angle
u 25.0°. (a) What is the magnitude of the
block’s acceleration? (b) The force magnitude F is
slowly increased. What is its value just before the
block is lifted (completely) off the floor? (c) What is
the magnitude of the block’s acceleration just before
it is lifted (completely) off the floor?

Answers

Answer 1
Answer: a.
the force in the direction of movement
Fx = F x cosA*
Fx = 12 x cos25*
Fx = 10.88
m x a = 10.88
a = 10.88/5
a= 2.18 m/s^2
b.
when the block will be lifted off the floor and when the vertical component
Fy = mg
Fy = F x sin25*
Fy= 5 x 9.8
Fy= 115.94 N
c.
if Fx = ma
a = 115.94/5
a= 23.19 m/s^2
hope it helps
Answer 2
Answer:

Final answer:

The magnitude of the block's acceleration is roughly 2.18 m/s². The force required just before the block is lifted off the floor is equal to the weight of the block, 49.05 N. Just before it is lifted, the block's acceleration is still due to the horizontal component of the force.

Explanation:

This problem involves physics concepts related to forces and acceleration of an object on a frictionless surface. The key to solving it involves understanding the relationship between mass, force, and acceleration (Newton's second law) and the concept of vertical and horizontal components of a force.

(a) The magnitude of the block's acceleration can be calculated using the horizontal component of the force (Fh = F cos θ) and Newton's second law (F = ma). So, a = Fh / m = (F cos θ) / m = (12.0 N cos 25) / 5.00 kg ≈ 2.18 m/s².

(b) The force F required to lift the block is equal to the weight of the block mg. Thus, F = m*g = 5.00 kg * 9.81 m/s² = 49.05 N.

(c) Just before the block is lifted off the floor, it is still in contact with the floor so the normal force is not zero. As such, there is no vertical acceleration, so the magnitude of the block's acceleration is still solely due to the horizontal component of F.

Learn more about Forces and Acceleration here:

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If a fully loaded shopping cart has a head-on collision with an empty cart both moving at the same speed toward each other, which cart experiences the greater force? The greater impulse? The greater change in momentum? The greater acceleration?

Answers

Answer:

impact- equal

impulse- equal

momentum- equal

acceleration- the empty cart

Can fertilizer grow plants (physics )

Answers

Answer:

A fertilizer or fertiliser is any material of natural or synthetic origin that is applied to soil or to plant tissues to supply one or more plant nutrients essential to the growth of plants. Many sources of fertilizer exist, both natural and industrially produced.

Explanation:

If 20 beats are produced within one second, which of the following frequencies could possibly be held by two sound waves traveling through a medium along a common path at the same time? A. 38 Hz and 63 Hz
B. 22 Hz and 42 Hz
C. 47 Hz and 55 Hz
D. 18 Hz and 26 Hz

Answers

Answer:

B. 22 Hz and 42 Hz

Explanation:

The beat is an interference pattern which occurs when two waves of slightly different frequencies meet each other. The beat is heard as a periodic variation, whose frequency (called beat frequency) is equal to the absolute value of the difference between the frequencies of the individual waves:

f_B = |f_1-f_2|

In this problem, we are told that 20 beats per second are produced, so the beat frequency is

f_B = 20 s^(-1) = 20 Hz

Therefore, we must find the two waves which have a difference in frequency equal to this value, 20 Hz. Let's see each choice:

A. 38 Hz and 63 Hz  --> f_B = |38 Hz-63 Hz|=25 Hz --> NO

B. 22 Hz and 42 Hz  --> f_B = |22 Hz-42 Hz|=20 Hz --> YES

C. 47 Hz and 55 Hz  --> f_B = |47 Hz-55 Hz|=8 Hz --> NO

D. 18 Hz and 26 Hz --> f_B = |18 Hz-26 Hz|=8 Hz --> NO

So, the correct choice is

B. 22 Hz and 42 Hz


The right answer for the question that is being asked and shown above is that: "B. 22 Hz and 42 Hz." If 20 beats are produced within one second, the frequencies that could possibly be held by two sound waves traveling through a medium along a common path at the same time is B. 22 Hz and 42 Hz 

PLEASE HELP 25 POINTS!!!! Multiply the following numbers and round your answer to the correct number of significant figures: 72,000 x 45

Show both the unrounded and rounded answer.

Answers

Answer:

3,240,000

Explanation:

72,000 times 45= 3,240,000

There is no rounded number unless u are dividing.

Answer:3240000
There is no rounding (dividing)

On a sky coaster (human pendulum) that reaches 20 meters from it's equilibrium position, a man of 70 kg is able to reach a maximum speed of

Answers

Answer:

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 20 m for h then

v=√(2*9.81*20)=19.799 m/s

Rounding off, v is approximately 20 m/s

Answer:

19.8 m/s

Explanation:

During the motion of a pemdulum bob, it casually converts kinetic energy to potential energy and vice versa.

A pendulum bob reaches its maximum speed at a position closest to its equilibrium position and has its lowest when it is farthest from the equilibrium position.

The maximum speed of a pendulum bob based on the mass involved and the maximum displacement from the equilibrium position is obtained from

Maximum kinetic energy = Maximum potential energy

Maximum potential energy occurs at the farthest point from equilibrium, that is,

P.E(max) = mgh

Maximum kinetic energy = ½mv²

½mv² = mgh

v = √2gh

g = acceleration due to gravity = 9.8 m/s²

h = farthest height from equilibrium position = 20 m

v = √(2×9.8×20) = 19.8 m/s

Hope this Helps!!!

A suitcase (mass m = 18 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures 0.50 m × 0.15 m. The elevator is moving upward with an acceleration of magnitude 1.6 m/s2. What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Answers

Answer:

P=2736 Pa

Explanation:

According to Newton we have that:

F=m*a\n

A force is exerted by the elevator to the suitcase, according to 3th Newton's law an equal force but in the opposite direction will appeared on the suitcase, that is:

F=m*g+m*a=m*(g+a)

F=205.2N

We know that the pressure is given by:

P=(F)/(A)\nP=(205.2N)/((0.50m)*(0.15m))\nP=2736N/m^2=2736Pa