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Elodia [21]
3 years ago
11

A force acts on a body of mass 13 kg initially at restThe force acts for 10 seconds, and once it quits, the body covers 60 m in

the next 6 seconds . Find the force that acted on the body
Physics
2 answers:
Ghella [55]3 years ago
6 0

Answer:

uguuffuufufufufucucucuvuvhvuchcyctxxtrxtccyyvuvubjbubbu

elixir [45]3 years ago
3 0

Answer: 13 N

Explanation: You have to use F=m * g

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Which layer of the atmosphere is directly above the troposhpere? A.troposhpere B.stratosphere C.mesosphere D.exoshpere
vivado [14]
Its B. i hope i helped!!!
4 0
3 years ago
Read 2 more answers
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0
3 years ago
Read 2 more answers
A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an
Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0

so we have

F_t = 129 N

\theta = 17.5^o

m = 38.2 kg

d = 85.4 m

so now we have

129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2

v = 8.57 m/s

7 0
3 years ago
List 5 reason why water is not thermometric liquid​
kolezko [41]

Answer:

(1) it is transparent so it makes the reading difficult. (2) it is volatile. (3) it is a poor conductor of heat. (4) it has a higher specific heat capacity, so it absorbs more heat from the body with which it is kept in contact. (5) Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids . If water is used in a thermometer , it will start phase change at 0oC and 100oC and will not measure temperature , out of this range .

PLEASE READ CAREFULLY AND PLEASE MARK AS BRAINLIEST :)

5 0
3 years ago
two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s
solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3)  of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

v3= 0.311 m/s

<h3>What is elastic collision velocity? </h3>
  • The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

#SPJ9

6 0
1 year ago
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