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Jlenok [28]
3 years ago
12

9) At a construction site, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened a

nd is half full, while the other box is new. The boxes, including the nails, weight 10 kg and 20 kg, respectively, and are the same size
a) As the plank tilts towards the heavier box, predict which box of nails will start to slide first. Explain your prediction
Physics
2 answers:
Varvara68 [4.7K]3 years ago
8 0

Answer:

The two boxes will start sliding at the same angle. Their different masses does not affect when one of the boxes will start sliding.

Explanation:

The two boxes will start sliding when the frictional force,Ff= The gravitational force,Fg

Ff= mgcos theta× coefficient of friction

Fg= mgsintheta

Ff=Fg= mgsintheta=mgcos theta×coefficient of friction

Dividing through with mg on both sides gives

Sintheta=ucostheta

Sintheta/costheta= Tan theta

Theta= tan^-1(coefficient of fricition)

From the solution mass of the boxes and gravity does not affect when the boxes will start sliding.

rosijanka [135]3 years ago
3 0
Box 1 (having a mass of 10 kg) will slide first as it has less inertia than the other box. It will also require less force to overcome static friction between the surface of the plank and that of the first box.
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A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the
Gelneren [198K]

Answer:

Option B. 2.8 s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 27 m/s

Angle of projection (θ) = 30

Acceleration due to gravity (g) = 9.8 m/s²

Time of flight (T) =?

The time of flight of the ball can be obtained as follow:

T = 2uSineθ / g

T = 2 × 27 × Sine 30 / 9.8

T = 2 × 27 × 0.5 / 9.8

T = 27 / 9.8

T = 2.8 s

Therefore, time of flight of the ball is 2.8 s

7 0
3 years ago
How do you find the number of neutrons
sammy [17]

the mass number minus the atomic number

8 0
3 years ago
Give two mathematical examples of Newton's third law and how you get the solution​
bagirrra123 [75]

Answer:

1) Any particle moving in a horizontal plane slowed by friction, deceleration = 32 μ

2) The particle moving by acceleration = P/m - 32μ OR The external force = ma + 32μm

Explanation:

* Lets revise Newton’s Third Law:

- For every action there is a reaction, equal in magnitude and opposite

 in direction.

- Examples:

# 1) A particle moving freely against friction in a horizontal plane

- When no external forces acts on the particle, then its equation of

  motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ No external force

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = 0 - F

∴ 0 - F = mass × acceleration

- Substitute F by μR

∴ - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ - μ(mg) = ma ⇒ a is the acceleration of motion

- By divide both sides by m

∴ - μ(g) = a

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ a = - 32 μ

* Any particle moving in a horizontal plane slowed by friction,

 deceleration = 32 μ

# 2) A particle moving under the action of an external force P in a

  horizontal plane.

- When an external force P acts on the particle, then its equation

 of motion is;

∵ ∑ forces in direction of motion = mass × acceleration

∵ The external force = P

∵ The friction force (F) = μR, where μ is coefficient of the frictional force

   and R is the normal reaction of the weight of the particle on the

   surface

∵ The frictional force is in opposite direction of the motion

∴ ∑ forces in the direction of motion = P - F

∴ P - F = mass × acceleration

- Substitute F by μR

∴ P - μR = mass × acceleration

∵ R = mg where m is the mass of the particle and g is the acceleration

  of gravity

∴ P - μ(mg) = ma ⇒ a is the acceleration of motion

∵ The acceleration of gravity ≅ 32 feet/sec²

∴ P - 32μm = ma ⇒ (1)

- divide both side by m

∴ a = (P - 32μm)/m ⇒ divide the 2 terms in the bracket by m

∴ a = P/m - 32μ

* The particle moving by acceleration = P/m - 32μ

- If you want to fin the external force P use equation (1)

∵ P - 32μm = ma ⇒ add 32μm to both sides

∴ P = ma + 32μm

* The external force = ma + 32μm

7 0
3 years ago
Which astronomer supported the belief that earth was at the center fo the universe?
MissTica
B. Ptolemy believed that the earth was the center of the universe
4 0
3 years ago
Read 2 more answers
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
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