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Annette [7]
3 years ago
13

An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 0.4 cm/s at an angle of 52◦ with the lon

g edge. How long will it take her to get across? Answer in units of s.
Physics
1 answer:
nordsb [41]3 years ago
5 0

Answer:

t=108.50s

Explanation:

The ant is under an uniform motion. So, we get the time taken by the ant to cross the strip from the definition of speed:

v=\frac{x}{t}\\t=\frac{x}{v}

In this case, we use the vertical speed of the ant (v_y=vsin\theta), since the ant crosses the strip that is 34.2 cm wide (vertically). So:

t=\frac{y}{v_y}\\t=\frac{y}{vsin\theta}\\t=\frac{34.2cm}{0.4\frac{cm}{s}sin(52^\circ)}\\t=108.50s

You might be interested in
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

6 0
3 years ago
Bartek w czasie 2minut przesunął ruchem jednostajnym kamień na odległość 0,5m.Oblicz moc mięśni Bartka jeśli siła oporów wynsiła
Ipatiy [6.2K]

Answer:

Bartek's muscle power = 2.667 W

Moc mięśni Bartka = 2.667 W

Explanation:

English Translation

Bartek within 2 minutes moved the stone

steadily at a distance of 0.5 m. Calculate Bartek's muscle power , if the resistance force was 640N.

The power, P, expended or generated by moving a body to move a velocity, v, against a resistive force, F is given as

P = Fv

P = ?

F = 640 N

v = velocity = (0.5) ÷ (2×60) = 0.00416667 m/s

P = 640 × 0.0041666667 = 2.667 W

In Polish/Po polsku

Moc P, zużyta lub wytworzona przez poruszanie ciałem w celu przemieszczenia prędkości v przeciwko sile oporowej, F jest podawana jako

P = Fv

P = ?

F = 640 N

v = velocity = (0.5) ÷ (2×60) = 0.00416667 m/s

P = 640 × 0.0041666667 = 2.667 W

Hope this Helps!!!

Mam nadzieję że to pomoże!!

4 0
3 years ago
Calculate the work done by Kevin as he lifts 3 boxes (mass=45 KG) 57 CM high with an acceleration recorded to be 12.3 m/s^2.
anyanavicka [17]
Answer = 31.5J (rounded to 1dp)

The question wants to find the work done, where work done can be calculated in the equation:

Work done = Force x Distance

The force is not given in the question, however they give the acceleration and mass, so you can calculate the force using Newton's second law:

Force = Mass x Acceleration

First, calculate the force. The mass is 45kg for the 3 boxes and the acceleration is 12.3m/s^2.

Force = 45 x 12.3

Force = 553.5N

Now that you have the force (553.5N) and distance (57cm), substitute those values into the equation for work done.

Make sure to convert the 57cm into metres (into SI units), otherwise the answer would be wrong.

57cm = 0.057m

Work done = Force x Distance

Work done = 553.5 x 0.057

Work done = 31.5J (rounded to 1dp)
8 0
2 years ago
Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end
ozzi

Answer:

The total work that the rope does to Mangnus is - 5780 Jules.

Explanation:

By definition, the work is defined as:

W=F.d

Where F and d are the force and the total displacement. Note that in the definition the product is a scalar product since F and d are both vectors.  

Take into account that according to third Newton's law the force that the rope does to Magnus is opposite to the force that Magnus does to the rope, therefore the scalar product will be negative due the rope's force goes against to Magnus displacement.  

For calculating the work, we take 2500 N as the value for the force and 2.312 meters as the value for the displacement:

W=-2500 N * 2.312 m

W=-5780 Nm = -5780 J

7 0
3 years ago
A potter's wheel is a uniform disk of mass of 10.0 kg and radius 20.0 cm. A 2.0-kg lump of clay, roughly cylindrical with radius
Lera25 [3.4K]

Answer:

b. 29.2 rev/min

Explanation:

  • Assuming no external torques acting during the process, total angular momentum must be conserved, as follows:

       L_{0} = L_{f}  (1)

  • The initial angular momentum L₀, can be expressed as follows:

        L_{0} = I_{0} * \omega_{0} (2)

        where I₀ = initial moment of inertia = moment of inertia of the disk +

        moment of inertia of the cylinder and ω₀ = initial angular velocity  =

       30.0 rev/min.

  • Replacing by the values, we get:I_{0} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{c} *r_{c} ^{2}  = 0.2 kg*m2 +9e-4 kg*m2 = 0.2009 kg*m2 (3)⇒ L₀ = I₀* ω₀ = 0.2009 kg*m² * 30.0 rev/min = 6.027 kg*m²*rev/min
  • The final angular momentum can be written as follows:

       L_{f} = I_{f} * \omega_{f} (4)

       where If = final moment of inertia = moment of the inertia of the solid

      disk + moment of  inertia of the clay flattened on a disk, and ωf = final

      angular velocity.

  • Replacing by the values, we get:

   I_{f} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{fd} *r_{fd} ^{2}  = 0.2 kg*m2 +6.4e-3 kg*m2 = 0.2064 kg*m2 (5)

       ⇒ Lo =Lf = If*ωf

  • Replacing (2) in (1), and solving for ωf, we get:

        \omega_{f} = \frac{L_{o}}{I_{f} } = \frac{6.027kg*m2*rev/min}{0.2064kg*m2} = 29.2 rev/min (6)

3 0
3 years ago
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