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larisa [96]
3 years ago
12

todos los cuerpos pueden estar sobre la tierra por la fuerza de gravedad .cuando nos caemos ,saltamos y realizamos otras activid

ades la fuerza de gravedad influye bastante ,por lo tanto ,esta fuerza la ejerce
Physics
1 answer:
const2013 [10]3 years ago
4 0
¿Es esta una pregunta verdadera o falsa? Por cierto, no hay mucha gente que hable español en esta aplicación, así que buena suerte.
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Can anyone solve this?
love history [14]

Answer:

F = 39.2 N

Explanation:

Since, the object is in uniform motion. Therefore, the frictional force on object will be:

Frictional Force = μk N = μk mg

where,

μk = coefficient of kinetic friction = 0.2

m = mass of crate = 10 kg

g = 9.8 m/s²

Therefore,

Frictional Force = (0.2)(10 kg)(9.8 m/s²)

Frictional Force = 19.6 N

The horizontal component of force must be equal to this frictional force to continue the uniform motion:

F Sin 30° = 19.6 N

F = 19.6 N/Sin 30°

<u>F = 39.2 N</u>

3 0
3 years ago
A object weighing 5 kg, starts to accelerate evenly on a horizontal line. A force moves the object
hammer [34]
Work= force*distance
Work= x*12
Force= mass*acceleration
Force= 5 kg*6
Force= 40 N
Work= 40×12
Work= 480 J (joules)
I think this is it
5 0
4 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
When developing an experimental design, which action would improve the
il63 [147K]

When developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.

<h3>What is experimental design?</h3>

Experimental design is a concept used to organize, conduct, and interpret results of experiments in an efficient way, making sure that as much useful information as possible is obtained by performing a small number of trials.

Thus, when developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.

Learn more about experimental design here: brainly.com/question/17274244

#SPJ1

8 0
2 years ago
An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?
NISA [10]

Answer:

Explanation:

Given parameters:

Weight of object  = 49N

Force applied = 12N

Unknown:

Acceleration of object  = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

 Acceleration  = \frac{12}{49}   = 0.25m/s²

3 0
3 years ago
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