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ANEK [815]
3 years ago
14

A block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane. (a) What is its velocity when it reaches t

he top of the plane? (b) How far horizontally does it land after it leaves the plane?
Physics
1 answer:
kramer3 years ago
4 0

Answer:

A.) 8 m/s

B.) 7.0 m

Explanation:

Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.

(a) What is its velocity when it reaches the top of the plane?

Since the plane is frictionless, the final velocity V will be the same as 8 m/s

The velocity will be 8 m/s as it reaches the top of the plane.

(b) How far horizontally does it land after it leaves the plane?

For frictionless plane,

a = gsinø

Acceleration a = 9.8sin28

Acceleration a = 4.6 m/s^2

Using the third equation of motion

V^2 = U^2 - 2as

Substitute the a and the U into the equation. Where V = 0

0 = 8^2 - 2 × 4.6 × S

9.2S = 64

S = 64/9.2

S = 6.956 m

S = 7.0 m

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Answer:

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Explanation:

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3 years ago
In order to control bacterial infections and/or food poisoning, the FDA (Food and Drug Administration) has approved the use of n
Lelechka [254]

The answer is food irradiation. This involves the brief exposure of food to gamma rays or X-rays to kill pathogens that may contribute to food spoilage. This increases the shelf-life of the food. Gamma rays and X-rays emanate from nuclear decay of radioactive materials such as uranium..

4 0
3 years ago
Read 2 more answers
Tâm ống dẫn đặt dưới đường phân giới giữa nước và thuỷ ngân h1 = 820mm, chênh lệch chiều cao cột thuỷ ngân h2 = 880mm (Hg = 13,
gayaneshka [121]

Answer:

P = 79993.43245

Explanation:

Được -

Vị trí tâm ống dẫn dưới đường thuỷ ngân h1 = 820 mm

Chênh lệch chiều cao của cột thủy ngân h2 = 880mm

Tỷ trọng của thủy ngân = 13,5951 g / cm3

P = mật độ * gia tốc do trọng lực a * h

Thay thế các giá trị đã cho, chúng ta nhận được -

P = 13,5951 g / cm3 * 980,665 cm / s2 * (88-82) cm

P = 79993.43245

8 0
3 years ago
Does anyone know this
NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net  force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the  right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

  • Force to the right: F_a = 20 N, the applied force
  • Force to the left: F_f = 4 N, the force of friction
  • Force to the bottom: F_g = 400 N, the weight of the box (the weight is always downward vertically)
  • Force to the top: F_N = 400 N. This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

F_x = F_a - F_f = 20 - 4 = 16 N (to the right)

While the vertical force is

F_y = F_N - F_g = 400 - 400 = 0

So the net force is 16 N to the right.

2)

In this case, we have the following forces:

  • F_g = 4 N downward, the weight of the ball
  • F_a = 100 N to the left, the force that kicks the ball
  • F_f = 2 N to the right, the force of friction
  • F_N = 4 N upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

F_x =F_a - F_f = 100 -2 = 98 N (to the left)

While the vertical force is

F_y = F_g - F_N = 4 - 4 = 0 (downward)

And so, the net force is 98 N to the left.

3)

The force acting on the squirrel in this problem are:

  • F_g = 8 N downward, the weight of the squirrel
  • F_f = 7.5 N upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

F_y = F_g - F_f = 8 - 7.5 = 0.5 N

And since the weight is larger than the air resistance, the direction of the net force is downward.

4)

The forces acting on Monkey are:

  • F_1=95 N is the force applied to the right by Bunny
  • F_2 = 75 N is the force applied by Deer from the left (so, also on the right)
  • F_g = 50 N is the weight of Monkey, downward
  • F_N = 50 N is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

F_x = F_1 + F_2 = 95+75=170 N (to the right)

While the net force in the vertical direction is

F_y = F_N - F_g = 50 - 50 = 0

And therefore the net force is 170 N to the right

5)

The forces acting on Deer are:

  • F_a = 100 N + 100 N = 200 N to the right, the combined force applied by Bunny and Monkey
  • F_f = 25 N to the left, the force of friction
  • F_g = 150 N downward, the weight of the deer
  • F_N = 150 N upward, the normal reaction from the surface that balances the weight

So the net horizontal force is

F_x = F_a - F_f = 200 - 25 = 175 N to the right

While the net vertical force is

F_y = F_N - F_g = 150 - 150 = 0

So the net force is 175 N to the right.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

3 0
3 years ago
Please help me with this question ASAP.
victus00 [196]

Answer:

i. 6.923 V

ii. The e.m.f. = 22.5 V

Explanation:

i. The given parameters are;

Length of potentiometer = 1 m

The resistance of the potentiometer = 10 Ω

The e. m. f. of the attached cell = 9 V

The current, I flowing in the circuit = e. m. f/(Total resistance)

The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A

The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire

The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V

ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

\dfrac{E}{R_{balance}} = \dfrac{V}{R_{cell}}

Where:

E = e.m.f. of the balance point cell

R_{balance} = Resistance of 75 cm of potentiometer wire  = 0.75×10 = 7.5 Ω

R_{cell} = Resistance of the cell in the circuit = 3 Ω

V = e.m.f. attached cell = 9 V

\dfrac{E}{7.5} = \dfrac{9}{3}

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V

7 0
3 years ago
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