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dybincka [34]
3 years ago
9

A physics student tests the theory of projectile motion by leaping off a 225 meter tall building. She runs off the building hori

zontally at 12.5 m/s. How far away from the base of the building should she place the safety net?
Physics
1 answer:
tamaranim1 [39]3 years ago
4 0
In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,

   d = V₀t + 0.5gt²

Substituting the known values,

  225 = (0 m/s)(t) + (0.5)(9.8)(t²)

Simplifying,
 
   t = 6.776 s

To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.


   range = (12.5 m/s)(6.776 s)

   range = 84.7 m

<em>Answer: 84.7 m</em>

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A 14.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay s
ahrayia [7]

Answer:

86.53 m/s

Explanation:

Given:

Mass of clay (m) = 14.0 g = 0.014 kg

Mass of block (M) = 110 g = 0.110 kg

Initial speed of block (U) = 0 m/s

Sliding distance (d) = 7.50 m

Coefficient of friction between block and surface (μ) = 0.650

Let the initial speed of clay be 'u' and speed of clay and block just after collision be 'v'.

Now, momentum is conserved just before and just after collision.

Momentum just before collision = mu + 0 = mu

Momentum just after collision = (m + M)v

Therefore, mu=(M+m)v --------- (1)

Now, using newton's second law and we find the acceleration of the system.

The frictional force is given as:

f=\mu mg=-ma\\\\a=-\mu g

Now, using equation of motion, we can find the velocity just after collision.

0^2=v^2+2ad\\\\v=\sqrt{-2ad}\\\\v=\sqrt{-2\times (-\mu g)\times d}\\\\v=\sqrt{2\mu gd}

Plug in the given values and find 'v'. This gives,

v=\sqrt{2\times 0.650\times 9.8\times 7.50}\\\\v=\sqrt{95.55}=9.77\ m/s

Now, using equation (1) and substituting the given values, we get:

0.014u=(0.014+0.110)\times 9.77\\\\u=\frac{0.124\times 9.77}{0.014}\\\\u=86.53\ m/s

Therefore, the speed of the clay immediately before impact is 86.53 m/s.

7 0
4 years ago
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velikii [3]
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8 0
4 years ago
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Bezzdna [24]
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A 45-mH inductor is connected in series with a 60-Ω resistor through a 15-V dc power supply and a switch. If the switch is close
egoroff_w [7]

Answer:

The current is 0.248 A

Explanation:

Given that,

Inductor L= 45\times10^{3}\ H

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We need to calculate the current

Using formula of current

I=\dfrac{V}{R}(1-e^{\dfrac{-R}{L}}t)

Where, V = voltage

R = resistance

L = inductance

T = time

Put the value into the formula

I=\dfrac{15}{60}(1-e^{\dfrac{-60}{45\times10^{3}}}\times7\times10^{-3})

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What other factor affected the gravitational force between the two objects in this simulation?
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The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases

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