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goblinko [34]
3 years ago
12

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

on between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg. (a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward
Physics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

mv_{1i}+Mv_{2i}=(m+M)v  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

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Answer:

\displaystyle w=3.478\ rad/sec

M=0.0182\ J

v=0.398\ m/s

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

\displaystyle w=\sqrt{\frac{g}{L}}

We have the length of the pendulum is L=0.81 meters, then we have

\displaystyle w=\sqrt{\frac{9.8}{0.81}}

\displaystyle w=3.478\ rad/sec

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

U=mgh

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

H+Y=L

And

H=L\ cos\alpha

Solving for Y

Y=L(1-cos\alpha )

Since\ \alpha=8.1^o, L=0.81\ m

Y=0.0081\ m

The potential energy is

U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)

U=0.0182\ J

The mechanical energy is, then

M=K+U=0+U=U

M=0.0182\ J

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

\displaystyle K=\frac{mv^2}{2}

Equating to the mechanical energy of the system (M)

\displaystyle \frac{mv^2}{2}=0.0182

Solving for v

\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}

v=0.398\ m/s

4 0
3 years ago
Which of the following is used in Sonata II's prepared piano?
Delvig [45]
It is C. Copper wire.
8 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

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Answer:

Too old(Ex. if real time is 1000 then they estimate >1000)

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Now nuclear test on planet have significant effect on temperature rise.

Also 14°C rise in temperature is good one because of this.

If future archaeologists only consider that  uniform function as above mentioned then they estimate more time then the real one.

Thus too old is right answer.

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Answer:

Honestly I cant understand that so please explain it someone

Explanation:

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