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Ket [755]
3 years ago
8

There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot fl

ying low and slow drops a weight; it takes 3.0 s to hit the ground, during which it travels a horizontal distance of 170 m. Now the pilot does a run at the same height but twice the speed. How much time does it take the weight to hit the ground?
Physics
1 answer:
nikdorinn [45]3 years ago
8 0

Answer:

Also 3s.

Explanation:

Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.

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In your own words, describe why melting ice with salt freezes cream. Compare your descriptions to your classmates and describe h
gizmo_the_mogwai [7]

Answer:

Water normally freezes at 0°C (32°F). Salt lowers the freezing temperature. (That is, it can remain a liquid at much lower temperatures.)

When sprinkled on ice, the salt lowers the freezing temperature of the water which effectively melts the ice when the salt dissolves into it. There is a limit to how low it can reduce the temperature, though. If the temperature drops below -9°C (15°F), it's too cold for the salt to dissolve into the ice.

When making ice cream, the salt lowers the temperature of the ice and water sufficiently enough to freeze the cream.

3 0
3 years ago
An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec
jolli1 [7]

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=H+\frac{v_o^2}{2g}

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

\displaystyle v_o=\sqrt{2gh_m}

v_o=\sqrt{2\cdot 9.8\cdot 77.5}

v_o=39\ m/s

(b)

The maximum time or the time taken by the object to reach its highest  point is calculated as follows:

\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

t_m=4\ s

7 0
3 years ago
5 minute deadline<br><br><br> List 2 academic and Scientific applications of radiation
Harrizon [31]
Academic- affect on substances, how it works
Scientific- helping cancer, making weapons
3 0
3 years ago
A rollercoaster car has 2500 J of potential energy and 160 J of
lara31 [8.8K]

Answer:

E_{k2}=2660 [J] kinetic energy.

Explanation:

The energy in the initial state i.e. when the rollercoaster is at the top is equal to the energy in the final state i.e. when it is at the bottom of the hill.

These states can be represented by means of the second equation.

E_{k1}+E_{p1}=E_{k2}\\160 + 2500 = E_{k2}\\E_{k2}=2660 [J]

Since the rollercoaster is located in the bottom of the hill where the potential energy level is zero, therefore there is only kinetic energy in the second state.

8 0
3 years ago
(a) and (b) ??Thanks <br>HELP <br>d8 =1​
garri49 [273]

Explanation:

(a) You can solve this using kinematics or energy.

Using kinematics:

a = F/m = 90 N / 4 kg = 22.5 m/s²

v₀ = 0 m/s

Δx = 5 m

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (22.5 m/s²) (5 m)

v = 15 m/s

Using energy:

W = ΔKE

Fd = ½ mv²

(90 N) (5 m) = ½ (4 kg) v²

v = 15 m/s

(b) ΔU = mg Δh

ΔU = (4 kg) (9.8 m/s) (12 m sin 40° − 15 m)

ΔU = -290 J

W = ΔU = -290 J

7 0
3 years ago
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