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BaLLatris [955]
3 years ago
12

A 55.2 kg softball player moving 3.11 m/s slides across dirt with uk=0.310. How far does she slide before coming to a stop​

Physics
1 answer:
hoa [83]3 years ago
5 0

Refer to the attachment

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A 90.00-kg hockey goalie, at rest in front of the goal, stops a puck (m = 0.16 kg) that is traveling at 30.00 m/s. at what speed
Ivahew [28]
<span>5.3 cm/s This is a matter of conservation of momentum. Since there's no mention of the puck rebounding, I will consider this to be a totally non-elastic collision. So, let's determine the starting momentum of the system. Goalie is at rest, so his momentum is 0. Puck is moving at 30.00 m/s with a mass of 0.16 kg, so: 30.00 m/s * 0.16 kg = 4.8 kg*m/s So the starting momentum is 4.8 kg*m/s moving towards the goal. After the collision, the puck and goalie will have the same momentum. So figure out the mass of the new system: 90.00 kg + 0.16 kg = 90.16 kg And divide the system momentum by the system mass: 4.8 kg*m/s / 90.16 kg = 0.053238687 m/s Finally, round to the least precise datum, so the result to 2 significant figures is 0.053 m/s, or 5.3 cm/s.</span>
6 0
4 years ago
Calcula la energia potencial de Tamara que tiene una mas de 50kg y se encuentra, 80 metros arriba en el borde de un edificio
Ilya [14]

Responder:

39200 J

Explicación:

Dado:

Masa de Tamara (m) = 50 kg

Altura a la que se encuentra Tamara (h) = 80 m

Aceleración debido a la gravedad (g) = 9.8 m / s²

La energía potencial de un objeto de masa 'm' ubicada a una altura 'h' sobre el suelo se da como:

U=mgh

Ahora, conecte los valores dados y resuelva la energía potencial. Esto da,

U=50\times 9.8\times 80\\\\U=39200\ J

Por lo tanto, la energía potencial de Tamara ubicada a una altura de 80 m es 39200 J.

8 0
4 years ago
ANSWER ASAP PLEASE!!
eimsori [14]
Sprry o cant see the words clearly
7 0
3 years ago
What is the maximum acceleration of a platform that oscillates at amplitude 4.70 cm and frequency 6.50 Hz?
mylen [45]

Answer:

78.315 m/s²

Explanation:

Amplitude, A = 4.70 cm = 0.047 m

Frequency, f = 6.50 Hz

Angular frequency, ω = 2 π f = 2 x 3.14 x 6.50 = 40.82 rad/s

Maximum acceleration, a = ω² A

a = 40.82 x 40.82 x 0.047

a = 78.315 m/s²

8 0
4 years ago
The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are
r-ruslan [8.4K]

Answer with explanation:

Part a)

v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s

Part b)

a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}

8 0
3 years ago
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