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VashaNatasha [74]

3 years ago

13

At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de

vice to stop the train so that it will not plow through the station if the engineer misjudges the stopping distance. While waiting, you wonder what would be the fastest train that the spring could stop at its full compression, 3.0 ft?

1 answer:

e-lub [12.9K]3 years ago

6 0

Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.

Answer:

2.99 m/s

Explanation:

Stopping distance, s = 3 ft = 0.914 m

final velocity, v = 0

a = g/2 = 4.9 m/s²

Use third equation of motion:

$v^2-u^2 = 2as$

substitute the values to find the speed of train:

$0 -u^2 = 2\times -4.9 \times 0.914 \\u^2=8.96 \\u=2.99 m/s$

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A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of h

mrs_skeptik [129]

Answer:

1199 miles

Explanation:

1 hour 30 minutes = 1 + 30/60 = 1.5 hours

2 hours 15 minutes = 2 + 15/60 = 2.25 hours

The distance she flew in the 1st segment is:

1.5*345 = 517.5 miles

The distance she flew in the 2nd segment is:

2.25 * 345 = 776.25 miles

Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown

776.25 * cos(45) = 549 miles in-line with the 1st segment and

776.25* sin(45) = 549 miles perpendicular to the 1st segment:

So the distance from the end to her starting position is

$\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles$

4 0

3 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

3 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago

A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.

Shalnov [3]

Answer:

Explanation:

During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s

v = u + at

1.2 = 0 + .8 a

a = 1.2 / .8

= 1.5 m /s²

During the acceleration in upward direction , let reaction force of ground on man be R .

Net force on man = R - mg

Applying Newton's 2 nd law

R - mg = ma

R = m ( g + a )

= 72 ( 9.8 + 1.5 )

= 813.6 N .

This reaction force will be measured by spring scale , so reading of spring scale will be 813.6 N .

3 0

2 years ago

A girl pulls on a 10-kg wagon with a constant horizontal force of 30 n. if there are no other horizontal forces, what is the wag

olga2289 [7]

Force = mass * acceleration
F = ma

Given m = 10 kg, F = 30 N;

F = ma
30 = 10a

Solving for a:
a = 3 m/s^2

The acceleration is 3 meters per second squared.

8 0

3 years ago