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3 years ago

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A 70-kg ice skater is standing still on the ice, when a friend tosses him a 10-kg ball that is traveling at +8 m/s. If he catche

s it, what would be the velocity of the person holding the ball? What type of collision is this?

1 answer:

Misha Larkins [42]3 years ago

7 0

Answer:+1.25 m/s

Explanation:

Given

mass of ice skater M=70 kg

mass of ball m=10 kg

the initial velocity of the ball $u_1=+8\ m/s$

Conserving linear momentum

$M\times0+m\timesu_1=(M+m)v\quad \quad [v=\text{combined velocity of skater and ball}]$

$v=\dfrac{10\times10}{80}=+1.25\ m/s$

Therefore the velocity of the person holding the ball is 1.25 m/s

This collision represents the perfectly inelastic collision where particles stick together after the collision.

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A stationary siren emits sound of frequency 1000 Hz and wavelength 0.343 m. An observer who is moving toward the siren will meas

Ainat [17]

Answer:

f>1000Hz and wavelength=0.343 m

Explanation:

We are given that

Frequency of stationary siren,f=1000 Hz

Wavelength of stationary sound, $\lambda=0.343 m$

When a observer is moving towards the siren then the frequency increases.

Therefore,an observer who is moving towards the siren measure a frequency >1000 Hz.

The wavelength depends upon the speed of source.

But we are given that siren is stationary.

Therefore, source is not moving and then the wavelength remains same.

f>1000Hz and wavelength=0.343 m

4 0

3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final:

$A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]$

$A_{f} = 5102.752\,cm^{2}$

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

4 0

3 years ago

Kenny klutz drops his physics book off his aunt's high-rise balcony.it hits the ground below 1.5 s later?

tatuchka [14]

Gravity causes a falling object to fall 9.8 m/s faster every second it falls.

Kenny's book started out with no speed when it was dropped.
1.5 sec later, it was falling at (9.8 x 1.5) = 14.7 m/s .

During the fall, its average speed was 1/2(0 + 14.7) = 7.35 m/s .

Distance it covered = (average speed) x (time) =

(7.35 m/s) x (1.5 sec) = 11.025 m

8 0

3 years ago

Need help! Photo says all! Will mark brainliest :)

AVprozaik [17]

Answer:

C-D

Explanation:

As you can see from the graph, the distance from A to B was from 0 m to 6 m in a duration of 3 seconds.

Divide 6 meters by 3 seconds to find the speed:

6 ÷ 3 = 2 m/s

B-C is not moving due to a straight line as said in the graph, so speed is

0 m/s.

There is also C-D since the car traveled from a distance of 9 meters

(6 -(-3) = 9) in 3 seconds too. (NOTE: The graph line going down does not mean it is slowing down, but rather going to a certain distance like going backwards)

Divide 9 meters by 3 seconds to get the speed:

9 ÷ 3 = 3 m/s

Between A-B, B-C, and C-D, C-D has the fastest speed recorded with 3 m/s.

A-D does not count here as the line has no connection between point A and point D.

Cheers!

6 0

3 years ago