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Anon25 [30]
3 years ago
6

Un tren viaja a 35 m/s y debe reducir su velocidad a 18 m/s al pasar por un puente. si tarda 5,5 segundos en reducir esta veloci

dad responde:
a) ¿que distancia recorre en este tiempo?
b) ¿cual es su aceleración?
Physics
1 answer:
a_sh-v [17]3 years ago
5 0

Kfzifxykzjdzriehdhaitzugthxjgare

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A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bott
Rina8888 [55]

Answer:

Explanation:

reading of scale = reaction force of surface R

centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .

R = mg + m v² / R

given ,

m v² / R = .80 mg

v² = .80 x g x R

= .8 x 9.8 x 9 = 70.56

v = 8.4 m /s

3 0
2 years ago
A clinical psychologist is one type of pure basic research psychologist . True are false
adoni [48]

false. clinical deals with patients and treats the.

research looks at root causes which clinical applies

8 0
3 years ago
The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont
Elza [17]

Answer:

1923 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

3 0
3 years ago
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
How much energy is required to heat 70 g of water at 20°C to boiling
choli [55]

Answer:

Q=23,430J

Explanation:

Hello,

In this case, since we compute the required energy via:

Q=mC\Delta T

Whereas m is the mass which here is 70 g, C the specific heat which for water is 4.184 J/(g°C) and ΔT is the temperature difference which is:

\Delta T=100-20=80\°C

Therefore, the energy turns out:

Q=70g*4.184\frac{J}{g\°C}*80\°C\\ \\Q=23,430J

Best regards.

3 0
3 years ago
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