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2 years ago

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In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start

s to careen down the steeply sloped track. In one event, the sled reaches a top speed of 9.2 m/s before starting down the initial part of the track, which is sloped downward at an angle of 4.0°. What is the sled's speed after it has traveled the first 140 m?

1 answer:

Hitman42 [59]2 years ago

4 0

Answer:

$v_f = 16.6 m/s$

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

$F_g = mg sin\theta$

so the acceleration due to gravity along the plane is given as

$a = \frac{F_g}{m}$

now we have

$a = g sin\theta$

$a = (9.81 sin4.0)$

$a = 0.68 m/s^2$

now we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 9.2^2 = 2(0.68)(140)$

$v_f = 16.6 m/s$

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You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi

aivan3 [116]

Answer:

$h=17005.8 km$

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass $M=6.39\times10^{23} kg$ at a distance r will be:

$F=\frac{GMm}{r^2}$

where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

$F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}$

Putting all together:

$\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}$

which means:

$r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}$

Which for our values is:

$r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km$

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

$h=r-R=20395.3km-3389.5 km=17005.8 km$

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2 years ago

A. Why is the stratosphere considered a "stable" layer in the atmosphere?

vovangra [49]

Answer:

A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.

B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.

C. The water vapor come from different kinds of fronts; gust fronts from existing storms as their downdraft hits the surface, spreads and lifts air in front, upper air disturbances and surface heating by solar radiation making an unbalanced vertical profile .

D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.

E. Wind shear is the change in wind direction or speed with height in an atmosphere.

Explanation:

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2 years ago

Susan and Hannah are each riding a swing. Susan has a mass of 25 kilograms, and Hannah has a mass of 30 kilograms. Susan’s swing

Charra [1.4K]

Answer:

Kinetic energy is given by:

K.E. = 0.5 m v²

Susan has mass, m = 25 kg

Velocity with which Susan moves is, v = 10 m/s

Hannah has mass, m' = 30 kg

Velocity with which Hannah moves is, v' = 8.5 m/s

<u>Kinetic energy of Susan:</u>

0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J

<u>Kinetic energy of Hannah:</u>

0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J

Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.

Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.

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2 years ago

(a) A small object with mass 3.75 kg moves counterclockwise with constant speed 1.55 rad/s in a circle of radius 2.55 m centered

Eddi Din [679]

Answer:3.95 m/s

Explanation:

Given

mass of object $m=3.75 kg$

$\omega =1.55 rad/s$

radius of circle $=2.55 m$

initial Position $r=2.55 \hat{i}$

angular displacement $\theta _0=8.95 rad$

8.95 radian can be written as

$1.42 (2\pi )$

i.e. Particle is at first quadrant with $\theta =0.4242\pi \times \frac{180}{\pi }$

$\theta =76.36^{\circ}$

(c)velocity is $v=\omgea \times r$

$v=1.55\times 2.55=3.95 m/s$

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2 years ago

A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When

kompoz [17]

Answer:

Total mass of ice = 38.06g

Explanation:

Since the heat capacity of calorimeter is negligible.

The water is already at 0°C, so the heat loss can no longer reduce the temperature of the water. It is used for fusion and forming more ice.

The equilibrium temperature will be 0°C, because the heat gain by ice is only enough to bring it down to 0°C.

Heat gained by ice = heat loss by water

Heat gained by ice (from -14°C to 0°C) = heat lost to fusion by water (heat of fusion of some amount of the water present in the calorimeter)

mi Ci ∆Ti = mw . L ......1

Where;

mi = mass of ice = 35g = 0.035 kg

Ci = specific heat capacity of ice = 2090 J/kg ∙ K

∆Ti = change in temperature of ice = 0-(-14) = 14 K

mw = the mass of water that have gained enough heat for fusion ( mass of water converted to ice)

L = latent heat of fusion of water = 33.5 × 10^4 J/kg.

From equation 1;

mw = (mi Ci ∆Ti )/L

mw = (0.035×2090×14)/335000

mw = 0.00306 kg

mw = 3.06 g

Therefore, 3.06 g of water has been converted to ice.

When combined with the initial amount of ice initially in the calorimeter (at 0°C)

Total mass of ice = mi + 3.06g = 35g + 3.06g = 38.06g

Total mass of ice = 38.06g

6 0

2 years ago

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