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cluponka [151]
3 years ago
13

A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are t

he upward forces?
Physics
1 answer:
mel-nik [20]3 years ago
5 0

If the book is placed in the middle, the forces acting on <em>p</em> and <em>q</em> is 5N. If the book is moved 50 cm from <em>q</em>, the forces at <em>p</em> and <em>q</em> can be solved by doing a moment balance

<u>With </u><u><em>p</em></u><u> as the pivot;</u>

Fq (2 m) = 10 N (0.5 m)

Fq = 2.5 N

Fp = 10 N - 2.5 N = 7.5 N

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A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1
tamaranim1 [39]
The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6
7 0
3 years ago
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A mass m is attached to a string connected to a force sensor on a rotating platform. The platform’s angular velocity, ω, can be
makkiz [27]

Answer:

The mass m is 0.332 kg or 332 gm

Explanation:

Given

The platform is rotating with angular speed , \omega =8.5\, \frac{rad}{sec}

Mass m is moving on platform in a circle with radius , r=0.20\, m

Force sensor reading to which spring is attached , F=4.8\, N

Now for the mass m to move in circle the required centripetal force is given by F=m\omega ^{2}r

=>4.8=m\times 8.5 ^{2}\times 0.20

=>m=0.332\, kg

Thus the mass m is 0.332 kg or 332 gm

7 0
3 years ago
What type of radioactive decay is shown in this equation?
svp [43]
There is no <span>radioactive decay</span>
6 0
3 years ago
You have a 16-kg suitcase. what work did you do when you have slowly lifted it 0.80 m upward?
Anna71 [15]
The formula is:
Work = Force · Displacement
F = m · g
F = 16 kg · 9.8 m/s²  = 156.8 N
and we know that:
d = 0.8 m
W = 156.8 N · 0.8 m = 125.44 J
Answer:
W = 125.44 J.
8 0
4 years ago
At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three different directions. As a res
hram777 [196]

Answer:

F₃ = 122.88 N

θ₃ = 20.63°

Explanation:

First we find the components of F₁:

For x-component:

F₁ₓ = F₁ Cos θ₁

F₁ₓ = (50 N) Cos 60°

F₁ₓ = 25 N

For y-component:

F₁y = F₁ Sin θ₁

F₁y = (50 N) Sin 60°

F₁y = 43.3 N

Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:

F₂ₓ = F₂ = 90 N

F₂y = 0 N

Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:

F₁ₓ + F₂ₓ + F₃ₓ = 0 N

25 N + 90 N + F₃ₓ = 0 N

F₃ₓ = - 115 N

for y-components:

F₁y + F₂y + F₃y = 0 N

43.3 N + 0 N + F₃y = 0 N

F₃y = - 43.3 N

Now, the magnitude of F₃ can be found as:

F₃ = √F₃ₓ² + F₃y²

F₃ = √[(- 115 N)² + (- 43.3 N)²]

<u>F₃ = 122.88 N</u>

and the direction is given as:

θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)

<u>θ₃ = 20.63°</u>

7 0
3 years ago
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