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3 years ago

An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?

Physics

2 answers:

anyanavicka [17]3 years ago

8 0

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

Papessa [141]3 years ago

3 0

Answer:

10 seconds

Explanation:

It takes 10 seconds for an object to fall from an elevation of 250 ft.

250 = 0 + (0) t + ½ (5) t²

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what is the focalength of the combination of two thin renses of power -5D and -2D placeci in contact with each other?

Ghella [55]

Answer:

f = - 0.143 m = - 14.3 cm

Explanation:

First, we will calculate the power of the combination of lenses:

P = P₁ + P₂

where,

P = Power of Combination = ?

P₁ = Power of first lens = - 5D

P₂ = Power of second lens = - 2D

Therefore,

P = - 5D - 2D

P = - 7D

Now, the focal length can be given as:

$f = \frac{1}{P} \\\\f = \frac{1}{-\ 7\ D} \\\\$

f = - 0.143 m = - 14.3 cm

Negative focal length indicates that combination will act as diverging lens.

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3 years ago

Given the quantities a= 9.7 m, b= 4.2 s, c= 69 m/s, what is the value of the quantity d = a^3/(cb^2)?

nexus9112 [7]

D= 9.7^3/(69)(4.2)^2
d=912.673/289.8^2
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4 years ago

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Can someone please help me with science.

pav-90 [236]

On the dog's return trip (between t = 10 and t = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.

You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is

(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1

while during the return trip, it is

(0 - 5) / (12.5 - 10) = -5/2.5 = -2

Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.

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3 years ago

Jake uses a fire extinguisher to put out a small fire. When he squeezes the handle, the flame rettardant is released from the ex

Tpy6a [65]

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4 years ago

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3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulle

Lena [83]

Answer:

$\mu=0.377$

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity ( $w_1$ and $w_2$ ) which equal the product of the block's mass times "g".

On the second block ( $m_2$ ), there are just two forces acting: the block's weight ( $m_2\,*\,g$ ) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

$x_f-x_i=v_i\,t-\frac{1}{2} a\,t^2$ and assume there was no initial velocity imparted to the block:

$x_f-x_i=v_i\,t-\frac{1}{2} a\,t^2\\-0.92\,m=0\,-\frac{1}{2} a\,(1.23)^2\\a=\frac{0.92\,*\,2}{1.23^2} \\a=1.216 \,\frac{m}{s^2}$

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

$F_{net}=m_2\,a\\w_2-T=m_2\,a\\m_2\,g-T=m_2\,a\\m_2\,g-m_2\,a=T\\m_2\,(g-a)=T\\1.2\,(9.8-1.216)\,N=T\\T=10.3008\,N$

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 $\frac{m}{s^2}$ , and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

$n-m_1\,g\,cos(12^o)=0\\n=m_1\,g\,cos(12^o)\\n=1.5\,*\,9.8\,cos(12^o)\\n=14.38\,N$

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration $a$ ), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (f ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

$F_{net}=m_1\,a\\T-f-w_1\,sin(12)=m_1\,a\\T-w_1\,sin(12)-m_1\,a=f\\f=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\\f=5.42\,N$

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown $\mu$ ) times the magnitude of the normal force (14.38 N):

$f=\mu\,n\\5.42\,N=\mu\,*\,14.38\,N\\\mu=\frac{5.42}{14.38}\\\mu=0.377$

with no units.

4 0

3 years ago