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olga nikolaevna [1]
3 years ago
12

A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the b

ottom moving along the ground when the bottom of the ladder is 5 ft from the wall
Physics
1 answer:
netineya [11]3 years ago
7 0

Answer: The bottom of the ladder is moving at 3.464ft/sec

Explanation:

The question defines a right angle triangle. Therefore using pythagorean

h^2 + l^2 = 10^2 = 100 ...eq1

dh/dt = -2ft/sec

dl/ dt = ?

Taking derivatives of time in eq 1 on both sides

2hdh/dt + 2ldl/dt = 0 ....eq2

Putting l = 5ft in eq2

h^ + 5^2 = 100

h^2 = 25 = 100

h Sqrt(75)

h = 8.66 ft

Put h = 8.66ft in eq2

2 × 8.66 × (-2) + 2 ×5 dl/dt

dl/dt = 17.32 / 5

dl/dt = 3.464ft/sec

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A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l
hram777 [196]

Answer:

t = 37.6 s

Explanation:

As we know that train is initially moving with the speed

v_i = 75.1 km/h

now we know that

v_i = 20.86 m/s

now the final speed of the train when it crossed the crossing

v_f = 15 km/h

v_f = 4.17 m/s

now we can use kinematics here

v_f^2 - v_i^2 = 2 a d

4.17^2 - 20.86^2 = 2 a(471)

a = -0.44 m/s^2

Now the time to cross that junction is given as

v_f - v_i = at

4.17 - 20.86 = a(-0.44)

t = 37.6 s

4 0
2 years ago
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0
2 years ago
Use the motion map to answer the question, describe the position and velocity of the object based moo the motion map?
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3 0
3 years ago
The wave that carrys the most energy are the waves that?
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5 0
2 years ago
Read 2 more answers
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p
beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0
3 years ago
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