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allochka39001 [22]
3 years ago
15

Big Al stands on a skateboard at a rest and throws a 0.5-kilogram rock at a velocity of 10.0 m /sec. Big Al moves back at 0.05 m

/sec. What is the combined mass of Big Al and the skateboard?
Physics
1 answer:
Vlada [557]3 years ago
4 0
Let
his and skateboard's combined mass is x
we know
F1=-F2
0.5kg*10m/s=-0.05m/s xkg
5=0.05x(minus cancel with the m/s as it represented the opposite direction of velocity and now there is no velocity in this equation.. so minus is avoidable. and the kgs cancel out)
x=10
so their combined mass is a 100 kg.. (I hope I didn't mess thing up for you)
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Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ T=2 \pi\sqrt{\frac{m}{k} }

or,

⇒ m=\frac{k T^2}{4 \pi^2}

On substituting the values, we get

⇒     =\frac{280\times (3)^2}{4 \pi^2}

⇒     =\frac{280\times 9}{4\times (3.14)^2}

⇒     =68.83 \ kg

(b)

The speed of the string will be:

⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

then,

⇒             v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }

On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

⇒                =0.725 \ m/s

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