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geniusboy [140]

3 years ago

6

Mary slides down a snow-covered hill on a large piece of cardboard and then slides across a frozen pond at a constant velocity o

f 2.40 m/s. After Mary has reached the bottom of the hill and is sliding across the ice, Sue runs after her at a velocity of 4.40 m/s and hops on the cardboard. How fast do the two of them slide across the ice together on the cardboard? Mary's mass is 69.0 kg and Sue's is 56.0 kg. Ignore the mass of the cardboard and any friction between the cardboard and the snow and/or ice. (Indicate the direction with the sign of your answer.) m/s

1 answer:

Tresset [83]3 years ago

3 0

Answer: 3.288 m/s

Explanation:

Given

Mass of Mary, m1 = 69 kg

Mass of Sue, m2 = 56 kg

Speed of Mary, v1 = 2.4 m/s

Speed of Sue, v2 = 4.4 m/s

Speed of the 2 of them, v = ?

We solve this using the principle of conservation of linear momentum

m1.v1 + m2.v2 = m1.v + m2.v

m1.v1 + m2.v2 = (m1 + m2) v

v = [(m1.v1) + (m2.v2)] / (m1 + m2)

v = [(69 * 2.4) + (56 * 4.4)] / (69 + 56)

v = (164.6 + 246.4) / 125

v = 411 / 125

v = 3.288 m/s

Thus, the speed at which both Mary and Sue slide together across the ice is 3.288 m/s

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A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

2 years ago

When devising a model, scientists can only use the information available during their lifetime. This means that the current mode

11111nata11111 [884]

no, it not useless. we still learn Bohr's model in HS n dats almost 200 yr old! while there may be new models, previous one is good for explaining the basics. it is also useful to learn previous model n see how our understanding improves over time.

6 0

3 years ago

Read 2 more answers

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a

zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

$\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s$

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

v = 8.19m/s

Thus, The tangential speed of the tack is 8.19 m/s.

4 0

3 years ago

Read 2 more answers

Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are

Juliette [100K]

Answer: $Q=248.011 W$

Explanation:

Given

Temperature of Room $T_{\infty }=18^{\circ}\approx 291 K$

Area of Person $A=1.7 m^2$

Temperature of skin $T=32^{\circ}\approx 305 K$

Heat transfer coefficient $h=5 W/m^2.k$

Emissivity of the skin and clothes $\epsilon =0.9$

$\Delta T=32-18=14^{\circ}$

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation $Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )$

where $\sigma =stefan-boltzman\ constant$

$Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)$

$Q_1=129.01 W$

Heat Transfer due to convection is given by

$Q_2=hA(\Delta T)$

$Q_2=5\times 1.7\times 14=119 W$

$Q=Q_1+Q_2$

$Q=129.01+119=248.011 W$

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3 years ago

A body is oscillating up and down at the end of a spring. Let’s consider when the body is at the top of its up-and-down motion.

Klio2033 [76]

The velocity of the body is zero; option A

<h3>What is the motion of an oscillating body?</h3>

The motion of an oscillating body is known as simple harmonic motion.

Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.

For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.

In conclusion, oscillating bodies undergo simple harmonic motion.

Learn more about simple harmonic motion at: brainly.com/question/24646514

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3 0

1 year ago