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posledela
3 years ago
12

Necesito ayuda urgente por favor, gracias 2) Sabiendo que el espacio que recorre un cuerpo en función del tiempo viene dado por

la siguiente ecuación: X(t)=t 2 -8t+6, donde x se mide en metros y t en segundos. a) Realizar la gráfica , posición-tiempo de en el intervalo [0,5] Nota: Tabular para t=0,1,2,3,4,5, ubicar los puntos sobre el plano y luego unir los puntos con una curva. b) Que distancia habrá recorrido en un minuto.
Physics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:b

Explanation:

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Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
Read 2 more answers
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t
Aleksandr-060686 [28]

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

B = \frac{\mu I}{2 \pi r}

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

7 0
3 years ago
Please help me with this
aev [14]

Answer:

the answer is UV Radiation

6 0
3 years ago
By what percent must one increase the tension in a guitar string to change the speed of waves on the string from 301 m/s to 343
garri49 [273]

Answer:

29.8 %

Explanation:

7 0
3 years ago
A child who is swimming toward shore at 0.78 m/s sees shark and picks up his speed
Gnoma [55]

Answer:

0.085m/s²

Explanation:

Use v²=v0²+2a(d)

solve for a

v²-v0²/2d=a

Plug in givens

1.89²-0.78²/2*17.5=a

Plug into calculator

a=0.085m/s²

3 0
3 years ago
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