Ask question

Ask question

All categories

3 years ago

13

Two acrobats flying through the air grab and hold onto each other in midair as part of a circus act.One acrobat has a mass of 60

kg and has a horizontal velocity of 5 m/s just before the grab.Another acrobat has a mass of 50 kg and has a horizontal velocity of -3 m/s just before the grab.Their horizontal velocity immediately after they grab onto each other is:

1 answer:

earnstyle [38]3 years ago

4 0

Answer:

1.36m/s

Explanation:

We are given that

Mass of one acrobat, $m_1=60 kg$

Mass of another acrobat, $m_2=50 kg$

$v_2=-3 m/s$

$v_1=5 m/s$

We have to find their velocity immediately after they grab onto each other.

The collision between two acrobat is inelastic

According to law of conservation of momentum

$m_1v_1+m_2v_2=(m_1+m_2)V$

Substitute the values

$60\times 5+50(-3)=(60+50)V$

$300-150=110V$

$V=\frac{300-150}{110}=1.36m/s$

You might be interested in

An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i

Diano4ka-milaya [45]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

First of all, we can use the lens equation to find the location of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q,

$\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}\\q=\frac{1}{0.047 cm^{-1}}=21.3 cm$

The distance of the image from the lens is positive, so we can already conclude that the image is real.

Now we can also write the magnification equation:

${h_i}=-h_o \frac{q}{p}$

where $h_i, h_o$ are the size of the image and of the object, respectively.

Substituting p = 50 cm and q = 21.3 cm, we have

${h_i}=-h_o \frac{21.3 cm}{50 cm}=-0.43 h_o$

So from this relationship we observe that:

$|h_i| < |h_o|$ --> this means that the image is smaller than the object, and

$h_i < 0$ --> this means that the image is inverted

so, the correct answer is

It is real, inverted, and smaller than the object.

4 0

3 years ago

A sensor is used to monitor the performance of a nuclear reactor. The sensor accu-rately reflects the state of the reactor with

Helen [10]

Answer:

The probability of an incorrect report is found to be 0.03 or 3%.

Explanation:

We will get an incorrect report in both the cases of false alarm or missing excessive radiation. Since, both are mutually exclusive events. Therefore, the probability of both events to occur simultaneously will be 0. Thus, the probability of an incorrect report will be the sum of the probability of false alarm and the probability of a missing radiation.

P (False Alarm) = 0.02

P (Missing Radiation) = 0.01

P(Incorrect Report) = P (False Alarm) + P(Missing Radiation)

P (Incorrect Report) = 0.02 +0.01

P(Incorrect Report) = 0.03 = 3%

7 0

3 years ago

The fourth harmonic on a string fixed at both ends shows

Charra [1.4K]

Answer:

I believe the answer is B) Two wavelengths

8 0

3 years ago

Ice has a specific heat of 2090 J/(kg C) and water has a specific heat of 4186 J/(kg C). Water has a latent heat of fusion of 3.

zepelin [54]

Answer:

3.1 × 10⁷ J

Explanation:

The total heat required is the sum of the heats required in each stage.

1) Solid: from -60°C to 0°C.

Q₁ = c(s) × m × ΔT = (2090 J/kg.°C) × 10 kg × (0°C - (-60°C)) = 1.3 × 10⁶ J

where,

c(s): specific heat of the solid

m: mass

ΔT: change in the temperature

2) Solid to liquid at 0°C

Q₂ = Qf × m = (3.3 × 10⁵ J/kg) × 10 kg = 3.3 × 10⁶ J

where,

Qf: latent heat of fusion

3) Liquid: from 0°C to 100°C

Q₃ = c(l) × m × ΔT = (4186J/kg.°C) × 10 kg × (100°C - 0°C) = 4.2 × 10⁶ J

where,

c(l): specific heat of the liquid

4) Liquid to gas at 100 °C

Q₄ = Qv × m = (2.26 × 10⁶ J/kg) × 10 kg = 2.26 × 10⁷ J

where,

Qv: latent heat of vaporization

Total heat

Q₁ + Q₂ + Q₃ + Q₄

1.3 × 10⁶ J + 3.3 × 10⁶ J + 4.2 × 10⁶ J + 2.26 × 10⁷ J = 3.1 × 10⁷ J

3 0

3 years ago

An ice cube of mass 50.0 gg can slide without friction up and down a 25.0 degreedegree slope. The ice cube is pressed against a

kompoz [17]

<em>We have assumed the distance to be 0.1 meters (not millimeters) since the question has issues when expressing the units</em>

Answer:

$d=0.60\ m$

Explanation:

<u>Energy Conversion</u>

We need to understand and apply the concepts of energy conversion to solve this problem. Three types of energy are manifested in the motion of the ice cube of mass m.

When it's above the ground level at a height h, it has potential gravitational energy, given by

$U=mgh$

If the cube is moving at speed v, it has kinetic energy, given by

$\displaystyle K=\frac{1}{2}mv^2$

Finally, when it compresses the spring, it has elastic energy:

$\displaystyle E=\frac{1}{2}kx^2$

Where x is the distance of compression and k is the spring constant

When the ice cube is released, it has potential gravitational energy which magnitude we cannot calculate since we don't have the height. Then it goes down the slope and acquires speed and kinetic energy until it stops when compressing the spring a distance x. In that very moment, the total energy is stored in its elastic form:

$\displaystyle E=\frac{1}{2}25\cdot 0.1^2=0.125\ J$

When the ice cube travels up powered by that energy, it has both kinetic and potential energies, and it stops up in the ramp and starts reversing direction when it runs out of speed, thus the total potential energy is

$mgh=0.125\ J$

Solving for h, knowing m=50 g=0.05 Kg

$\displaystyle h=\frac{0.125}{mg}=\frac{0.125}{0.05\cdot 9.8}=0.26\ m$

The height and the distance traveled in the slope d are related by

$h=d.sin25^o$

Thus

$\displaystyle d=\frac{h}{sin25^o}=\frac{0.26}{sin25^o}=0.60\ m$

$\boxed {d=0.60\ m}$

4 0

4 years ago